ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

More Exercises

Question 1.
Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
⇒ y – √3x – 4

Question 2.
Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
Slope of the line 3y + 2x = 12
⇒ 3y = 12 – 2x
⇒ 3y = -2x + 12

Question 3.
If the equation of a line is y – √3x + 1, find its inclination.
Solution:
In the line
y = √3 x + 1
Slope = √3
⇒ tan θ = √3
⇒ θ = 60° (∵ tan 60° = √3)

Question 4.
If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.
Solution:
The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points -4 = 2m + c …(i)
and 1 = -3m + c …(ii)
Subtracting we get,

Question 5.
If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.
Solution:
Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)
divides AC in the ratio of m1 : m2

Question 6.
Find the inclination of the line joining the points P (4, 0) and Q (7, 3).
Solution:
Slope of the line joining the points P (4, 0) and Q (7, 3)

Question 7.
Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to $$– \frac { 3 }{ 7 }$$
Solution:
Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5

Question 8.
If point A is reflected in the y-axis, the co-ordinates of its image A1, are (4, – 3),
(i) Find the co-ordinates of A
(ii) Find the co-ordinates of A2, A3 the images of the points A, A1, Respectively under reflection in the line x = – 2
Solution:
(i) ∵ A is reflected in the y-axis and its image is A1 (4, -3)
Co-ordinates of A will be (-4, -3)

Question 9.
If the lines $$\frac { x }{ 3 } +\frac { y }{ 4 } =7$$ and 3x + ky = 11 are perpendicular to each other, find the value of k.
Solution:
Given Equation of lines are

Question 10.
Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).
Solution:
The equation of the line is x – 2y + 8 = 0
⇒ 2y = x + 8

Question 11.
Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.
Solution:
Equations of the line is
3y = 5 – x ⇒ 3y = -x + 5

Question 12.
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Solution:
Let slope of the line joining the points A (1, 2) and B (6, 7) be m1

Question 13.
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Solution:
Slope of line AC (m1)

Question 14.
A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:

(i) the co-ordinates of A and B
(ii) the equation of the line AB
Solution:
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (2, 1) divides BA in the ratio 3 : 1.

Question 15.
A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.
Solution:
Let the line make intercept a and b with the
x-axis and y-axis respectively then the line passes through

Question 16.
If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.
Solution:
Co-ordinates of A are (3, -2).

Diagonals AC and BD of the square ABCD
bisect each other at right angle at O.
∴ O is the mid-point of AC and BD Equation

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS

More Exercises

Choose the correct answer from the given four options (1 to 13) :

Question 1.
The slope of a line parallel to y-axis is
(a) 0
(b) 1
(c) – 1
(d) not defined
Solution:
Slope of a line parallel to y-axis is not defined. (b)

Question 2.
The slope of a line which makes an angle of 30° with the positive direction of x-axis is
(a) 1
(b) $$\frac { 1 }{ \sqrt { 3 } }$$
(c) √3
(d) $$– \frac { 1 }{ \sqrt { 3 } }$$
Solution:
Slope of a line which makes an angle of 30°
with positive direction of x-axis = tan 30°
= $$\frac { 1 }{ \sqrt { 3 } }$$ (b)

Question 3.
The slope of the line passing through the points (0, – 4) and ( – 6, 2) is
(a) 0
(b) 1
(c) – 1
(d) 6
Solution:
Slope of the line passing through the points (0, -4) and (-6, 2)
$$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2+4 }{ -6-0 } =\frac { 6 }{ -6 } =-1$$ (c)

Question 4.
The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is
(a) 0
(b) 1
(c) $$– \frac { 1 }{ 10 }$$
(d) not defined
Solution:
Slope of the line passing through the points (3, -2) and (-7, -2)
$$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ -7-3 } =\frac { 0 }{ -10 } =0$$ (a)

Question 5.
The slope of the fine passing through the points (3, – 2) and (3, – 4) is
(a) – 2
(b) 0
(c) 1
(d) not defined
Solution:
The slope of the line passing through (3, -2) and (3, -4)
$$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -4+2 }{ 3-3 } =\frac { -2 }{ 0 }$$ (d)

Question 6.
The inclination of the line y = √3x – 5 is
(a) 30°
(b) 60°
(c) 45°
(d) 0°
Solution:
The inclination of the line y = √3x – 5 is
√3 = tan 60° = 60° (b)

Question 7.
If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is
(a) -2
(b) -1
(c) 1
(d) 2
Solution:
Slope of the line passing through the points (2, 5) and (k, 3) is 2, then

Question 8.
The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is
(a) $$– \frac { 1 }{ 5 }$$
(b) $$\\ \frac { 1 }{ 5 }$$
(c) -5
(d) 5
Solution:
Slope of the line parallel to the line passing through (0, 6) and (7, 3)
Slope of the line = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3-6 }{ 7-0 } =\frac { -3 }{ 7 }$$ (b)

Question 9.
The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is
(a) $$– \frac { 1 }{ 5 }$$
(b) $$\\ \frac { 1 }{ 5 }$$
(c) -5
(d) 5
Solution:
Slope of the line joining the points (2, 5), (-3, 6)

Question 10.
The slope of a line parallel to the line 2x + 3y – 7 = 0 is
(a) $$– \frac { 2 }{ 3 }$$
(b) $$\\ \frac { 2 }{ 3 }$$
(c) $$– \frac { 3 }{ 2 }$$
(d) $$\\ \frac { 3 }{ 2 }$$
Solution:
The slope of a line parallel to the line 2x + 3y – 7 = 0
slope of the line

Question 11.
The slope of a line perpendicular to the line 3x = 4y + 11 is
(a) $$\\ \frac { 3 }{ 4 }$$
(b) $$– \frac { 3 }{ 4 }$$
(c) $$\\ \frac { 4 }{ 3 }$$
(d) $$– \frac { 4 }{ 3 }$$
Solution:
slope of a line perpendicular to the line 3x = 4y + 11 is

Question 12.
If the lines 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is
(a) 4
(b) – 4
(c) $$\\ \frac { 1 }{ 4 }$$
(d) $$– \frac { 1 }{ 4 }$$
Solution:
lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5

Question 13.
If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is
(a) $$\\ \frac { 3 }{ 2 }$$
(b) $$– \frac { 3 }{ 2 }$$
(c) $$\\ \frac { 2 }{ 3 }$$
(d) $$– \frac { 2 }{ 3 }$$
Solution:
line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0
are perpendicular to each other

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2

More Exercises

Question 1.
State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are
(i) parallel
(ii) perpendicular
(iii) neither parallel nor perpendicular.
Solution:
Slope of line y = 3x – 5 = 3
and slope of line 2y = 4x + 7
⇒ y = 2x + $$\\ \frac { 7 }{ 2 }$$ = 2.
∴ Slope of both the lines are neither equal nor their product is – 1.
∴ These line are neither parallel nor perpendicular.

Worksheets for Class 10 Maths

Question 2.
If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.
Solution:
In equation
6x + 5 y – 7 = 0
⇒ 5y = -6x + 7

Question 3.
Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b. (1991)
Solution:
In equation 2x – by + 5 = 0
⇒ – by = – 2x – 5
⇒ y = $$\\ \frac { 2 }{ b }$$ + $$\\ \frac { 5 }{ b }$$

Question 4.
Given that the line $$\\ \frac { y }{ 2 }$$ = x – p and the line ax + 5 = 3y are parallel, find the value of a. (1992)
Solution:
In equation y = x – p
⇒ y = 2x – 2p
Slope (m1) = 2
In equation ax + 5 = 3y

Question 5.
If the lines y = 3x + 7 and 2y + px = 3 perpendicular to each other, find the value of p. (2006)
Solution:
Gradient m1 of the line y = 3x + 7 is 3
2y + px = 3

Question 6.
Find the value of k for which the lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. (2003)
Solution:
Given
In equation, kx – 5y + 4 = 0

Question 7.
If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.
Solution:
Given
In the equation 3x + by + 5 = 0

Question 8.
Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?
Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1). (1993)
Solution:
Slope of the line passing through the points
(-2, 3) and (4, 1) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$$

Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (4, 1).

Question 9.
The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
Gradient (m1) of the line passing through the
points A (-2, 3) and B (4, b)

Question 10.
If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.
Solution:
In the line 3x + y = 4 …(i)
⇒ y = – 3x + 4
Slope (m1) = – 3
In the line x – ay + 7 = 0…..(ii)

⇒ -b = -6 ⇒ b = 6
Hence a = 3, b = 6

Question 11.
Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis. (1998)
Solution:
In the given line 2x – 3y – 7 = 0
⇒ 3y = 2x – 7
⇒ $$y=\frac { 2 }{ 3 } x-\frac { 7 }{ 3 }$$

Question 12.
Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.
Solution:
In the line 2x + 5y + 7 = 0
⇒ 5y = – 2x – 7
⇒ $$y=\frac { -2 }{ 5 } x-\frac { 7 }{ 5 }$$

Question 13.
Find the equation of a st. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.
Solution:
In the given line 3x – 4y + 12 = 0
⇒ 4y = 3x + 12
⇒ y = $$\\ \frac { 3 }{ 4 } x+3$$

Question 14.
Find the equation of the line which is parallel to 3x – 2y = – 4 and passes through the point (0, 3). (1990)
Solution:
In the given line 3x – 2y = – 4
⇒ 2y = 3x + 4
⇒ y = $$\\ \frac { 3 }{ 2 } x+2$$

Question 15.
Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0. (1999)
Solution:
In the given equation 3x + 5y + 15 = 0
⇒ 5y = – 3x – 15
⇒ y = $$\\ \frac { -3 }{ 5 } x-3$$

Question 16.
The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0, 5).
Solution:
In the given line y = 3x – 5
Here slope (m1) = 3
Substituting x = 0, then y = – 5
y-intercept = – 5
The slope of the line parallel to the given line
will be 3 and passes through the point (0, 5).
Equation of the line will be

Question 17.
Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point ( – 1, – 2).
Solution:
In the given line 3x + 8y = 12
⇒ 8y = -3x + 12
⇒ $$y=\frac { -3 }{ 8 } x+\frac { 12 }{ 8 }$$

Question 18.
(i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0. (1993)
Solution:
(i) In the line 4x – 3y + 12 = 0 …(i)
3y = 4x + 12
⇒ y = $$\\ \frac { 4 }{ 3 } x+4$$

Question 19.
Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and ( – 4, 1).
Solution:
The given line 2x + 5y – 7 = 0
5y = -2x + 7
⇒ $$y=\frac { -2 }{ 5 } x+\frac { 7 }{ 5 }$$

Question 20.
Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2), (2, 2).
Solution:
In the given line 3x + 2y – 8=0
⇒ 2y = – 3x + 8
⇒ y = $$\\ \frac { -3 }{ 2 } x+4$$

Question 21.
Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.
Solution:
Let the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)
Substituting the value of y in the equation
2x + 5 × 0 – 4 = 0
⇒ 2x – 4 = 0
⇒ 2x = 4
⇒ x = $$\\ \frac { 4 }{ 2 }$$ = 2
Coordinates of the points of intersection will be (2, 0)

Question 22.
The equation of a line is 3x + 4y – 7 = 0. Find
(i) the slope of the line. .
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0. (2010)
Solution:
(i) Equation of the line is 3x + 4y – 1 = 0
⇒ 4y = 7 – 3x

Question 23.
Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.
Solution:
In the equation 4x – 3y – 5 = 0,
⇒ 3y = 4x – 5
⇒ $$y=\frac { 4 }{ 3 } x-\frac { 5 }{ 3 }$$

Question 24.
Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).
Solution:
Given that
Slope of the line through (0, 0) and (2, 3)

Question 25.
Prove that the line through,( – 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).
Solution:
Given that
Slope of the line through (-2, 6) and (4, 8)

Question 26.
Show that the triangle formed by the points A (1, 3), B (3, – 1) and C ( – 5, – 5) is a right angled triangle by using slopes.
Solution:
Slope (m1) of line by joining the points
A(1, 3), B (3, -1) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } }$$

Question 27.
Find the equation of the line through the point ( – 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).
Solution:
Slope of the line joining the points (0, -2) and (4, 5) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } }$$
= $$\frac { 5+2 }{ 4-0 } =\frac { 7 }{ 4 }$$
Slope of the line parallel to it passing through (-1, 3) = $$\\ \frac { 7 }{ 4 }$$

Question 28.
A ( – 1, 3), B (4, 2), C (3, – 2) are the vertices of a triangle.
(i) Find the coordinates of the centroid G of the triangle.
(ii) Find the equation of the line through G and parallel to AC
Solution:
Given, A (-1, 3), B (4, 2), C (3, -2)
(i) Coordinates of centroid G

Question 29.
The line through P (5, 3) intersects y-axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the coordinates of Q.

Solution:
(i) Here θ = 45°
So, slope of the line = tan θ = tan 45° = 1
(ii) Equation of the line through P and Q is
y – 3 = 1(x – 5) ⇒ y – x + 2 = 0
(iii) Let the coordinates of Q be (0, y)

Question 30.
In the adjoining diagram, write down
(i) the co-ordinates of the points A, B and C.
(ii) the equation of the line through A parallel to BC. (2005)

Solution:
From the given figure, it is clear that
co-ordinates of A are (2, 3) of B are ( -1, 2)
and of C are (3, 0).

Question 31.
Find the equation of the line through (0, – 3) and perpendicular to the line joining the points ( – 3, 2) and (9, 1).
Solution:
The slope (m1) of the line joining the points (-3, 2) and (9, 1)

Question 32.
The vertices of a triangle are A (10, 4), B (4, – 9) and C ( – 2, – 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.
Solution:
Vertices of ∆ABC are A (10, 4), B (4, -9) and C( – 2, – 1)
Slope of the line BC (m1) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$

Question 33.
A (2, – 4), B (3, 3) and C ( – 1, 5) are the vertices of triangle ABC. Find the equation of :
(i) the median of the triangle through A
(ii) the altitude of the triangle through B
Solution:
(i) D is the mid-point of BC
Co-ordinates of D will

Question 34.
Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).
Solution:
Slope of the line joining the points (1, 2) and (5, -6)
$${ m }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -6-2 }{ 5-1 } =\frac { -8 }{ 4 } =-2$$

Question 35.
Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ if ( – 2, p) lies on it.
Solution:
Coordinates of A are (7, -3), of B = (1, 9)
(i) ∴ slope (m)

Question 36.
The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.
Solution:
Slope of BD (m1) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
= $$\frac { 8-3 }{ 6-1 } =\frac { 5 }{ 5 } =1$$
Diagonal AC is perpendicular bisector of diagonal BD

Question 37.
ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD. (2000)
Solution:
Co-ordinates of A (3, 6), C (-1, 2)
Slope of AC (m1) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
= $$\frac { 2-6 }{ -1-3 } =\frac { -4 }{ -4 } =1$$

Question 38.
Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and
(i) parallel to the line x + 2y – 5 = 0
(ii) perpendicular to the x-axis.
Solution:
4x + 3y = 1 …(i)
5x + 4y = 2 …(ii)
Multiplying (i) by 4 and (ii) by 3

Question 39.
(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP where 0 is the origin
(iii) In what ratio does the y-axis divide the line AB?
Solution:
(i) Co-ordinate A (-4, 1) and B (17, 10) P divides it in the ratio of 1 : 2
Let the co-ordinates of P will be (x, y)

Question 40.
Find the image of the point (1, 2) in the line x – 2y – 7 = 0
Solution:
Draw a perpendicular from the point P(1, 2) on the line, x – 2y – 7 = 0
Let P’ is the image of P and let its
co-ordinates sue (α, β) slope of line x – 2y – 7 = 0

Question 41.
If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.
Solution:
Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the

perpendicular bisector of PQ and it intersects the line at M.
M is the mid point of PQ Now slope of line x – 4y – 6 = 0

Question 42.
OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.
Solution:
OA = $$\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } }$$
$$\sqrt { { (3-0) }^{ 2 }+{ (0) }^{ 2 } }$$

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1

More Exercises

Question 1.
Find the slope of a line whose inclination is
(i) 45°
(ii) 30°
Solution:
(i) tan 45° = 1
(ii) tan 30° = $$\frac { 1 }{ \sqrt { 3 } }$$

Question 2.
Find the inclination of a line whose gradient is
(i) 1
(ii) √3
(iii) $$\frac { 1 }{ \sqrt { 3 } }$$
Solution:
(i) tan θ = 1 ⇒ θ = 45°
(ii) tan θ = √3 ⇒ θ = 60°
(iii) tan θ = $$\frac { 1 }{ \sqrt { 3 } }$$ ⇒ θ = 30°

Question 3.
Find the equation of a straight line parallel 1 to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it.
Solution:
(i) A line which is parallel to x-axis is y = a
⇒ y = 2
⇒ y – 2 = 0
(ii) A line which is parallel to x-axis is y = a
⇒ y = -3
⇒ y + 3 = 0

Question 4.
Find the equation of a straight line parallel to y-axis which is at a distance of:
(i) 3 units to the right
(ii) 2 units to the left.
Solution:
(i) The equation of line parallel to y-axis is at a distance of 3 units to the right is x = 3 ⇒ x – 3 = 0
(ii) The equation of line parallel to y-axis at a distance of 2 units to the left is x = -2 ⇒ x + 2 = 0

Question 5.
Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).
Solution:
The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3
⇒ x + 3 = 0

Question 6.
Find the equation of the a line whose
(i) slope = 3, y-intercept = – 5
(ii) slope = $$– \frac { 2 }{ 7 }$$, y-intercept = 3
(iii) gradient = √3, y-intercept = $$– \frac { 4 }{ 3 }$$
(iv) inclination = 30°,y-intercept = 2
Solution:
Equation of a line whose slope and y-intercept is given is
y = mx + c
where m is the slope and c is the y-intercept
(i) y = mx + c
⇒ y = 3x + (-5)
⇒ y = 3x – 5

Question 7.
Find the slope and y-intercept of the following lines:
(i) x – 2y – 1 = 0
(ii) 4x – 5y – 9 = – 0
(iii) 3x +5y + 7 = 0
(iv) $$\frac { x }{ 3 } +\frac { y }{ 4 } =1$$
(v) y – 3 = 0
(vi) x – 3 = 0
Solution:
We know that in the equation
y = mx + c, m is the slope and c is the y-intercept.
Now using this, we find,

Question 8.
The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.
Solution:
Equation of line PQ is 3y – 3x + 7 = 0
Writing in form of y = mx + c

Question 9.
The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.

Solution:
Slope of the line y = x + 1 after comparing
it with y = mx + c, m = 1

Question 10.
Find the value of p, given that the line $$\frac { y }{ 2 } =x-p$$ passes through the point ( – 4, 4) (1992).
Solution:
Equation of line is $$\frac { y }{ 2 } =x-p$$
It passes through the points (-4, 4)
It will satisfy the equation

Question 11.
Given that (a, 2a) lies on the line $$\frac { y }{ 2 } =3x-6$$ find the value of a.
Solution:
∵ Point (a, 2a) lies on the line
$$\frac { y }{ 2 } =3x-6$$
∴this point will satisfy the equation

Question 12.
The graph of the equation y = mx + c passes through the points (1, 4) and ( – 2, – 5). Determine the values of m and c.
Solution:
Equation of the line is y = mx + c
∴ it passes through the points (1, 4)
∴ 4 = m x 1 + c
⇒ 4 = m + c
⇒ m + c = 4 … (i)
Again it passes through the point (-2, -5)
∴ 5 = m (-2) + c
⇒ 5 = -2 m + c
⇒ 2m – c = 5 …(ii)
3m = 9
⇒ m = 3
Substituting the value of m in (i)
3 + c = 4
⇒ c = 4 – 3 = 1
Hence m = 3, c = 1

Question 13.
Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.
Solution:
∴ The line intersects y-axis making an intercept of -3
∴ the co-ordinates of point of intersection will be (0, -3)
Now the slope of line (m) = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$

Question 14.
Find the equation of a straight line passing through ( – 1, 2) and whose slope is $$\\ \frac { 2 }{ 5 }$$.
Solution:
Equation of the line will be
$$y-{ y }_{ 1 }=m(x-{ x }_{ 1 })$$
$$y-2=\frac { 2 }{ 5 } (x+1)$$
⇒ 5y – 10 = 2x + 2
⇒ 2x – 5y + 2 + 10 = 0
⇒ 2x – 5y + 12 = 0

Question 15.
Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).
Solution:
The equation of line whose slope is wand passes through a given point is
$$y-{ y }_{ 1 }=m(x-{ x }_{ 1 })$$
Here m = tan 60° = √3 and point is (0, -3)
∴ y + 3 = √3 (x – 0)
⇒ y + 3 = √3x
⇒ √3x – y – 3 = 0

Question 16.
Find the gradient of a line passing through the following pairs of points.
(i) (0, – 2), (3, 4)
(ii) (3, – 7), ( – 1, 8)
Solution:
m = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
Given
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)

Question 17.
The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :
(ii) The equation of EF
(iii) The coordinates of the point where the line EF intersects the x-axis.
Solution:
Co-ordinates of points E (0, 4) and F (3, 7) are given, then

Question 18.
Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.
Solution:
Putting y = 0, we will get the intercept made on x-axis,
2x – 3y + 12 = 0
⇒ 2x – 3 × 0 + 12 = 0
⇒ 2x – 0 + 2 = 0
⇒ 2x = -12
⇒ x = -6
and putting x = 0, we get the intercepts made on y-axis,
2x – 3y + 12 = 0
⇒ 2 × 0 – 3y + 12 = 0
⇒ -3y = -12
⇒ $$y= \frac { -12 }{ -3 }$$ = 4

Question 19.
Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.
Solution:
The two given points are P (5, 1), Q(1, -1).
∴ Slope of the line (m)

Question 20.
Find the value of ‘a’ for which the following points A (a, 3), B (2,1) and C (5, a) are collinear. Hence find the equation of the line.
Solution:
Given That
A(a, 3), B (2, 1) and C (5, a) are collinear.
Slope of AB = Slope of BC

Question 21.
Use a graph paper for this question. The graph of a linear equation in x and y, passes through A ( – 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (½, k). (2005)
Solution:
Points (h, 4) and (½, k) lie on the line passing
through A(-1, -1) and B(2, 5)

Question 22.
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find
(i) the coordinates of A
(ii) the equation of the diagonal BD.
Solution:
Given that
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)

Question 23.
In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.
Solution:
⇒ D is mid point of BC
∴ D is $$\left( \frac { 7+1 }{ 2 } ,\frac { 8-10 }{ 2 } \right)$$
i.e (4, -1)

Question 24.
Find the equation of a line passing through the point ( – 2, 3) and having x-intercept 4 units. (2002)
Solution:
x-intercept = 4
∴ Co-ordinates of the point will be (4, 0)
Now slope of the line passing through the points (-2, 3) and (4, 0)

Question 25.
Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.
Solution:
x-intercept = 6
∴ The line will pass through the point (6, 0)
y -intercept = -4 ⇒ c = -4

Question 26.
Write down the equation of the line whose gradient is $$\\ \frac { 1 }{ 2 }$$ and which passes through P where P divides the line segment joining A ( – 2, 6) and B (3, – 4) in the ratio 2 : 3. (2001)
Solution:
P divides the line segment joining the points
A (-2, 6) and (3, -4) in the ratio 2 : 3
∴ Co-ordinates of P will be

Question 27.
Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.
Solution:
line x – 2y – 11 = 0 passes through y-axis
x = 0,
Now substituting the value of x in the equation x – 2y – 11 = 0

Question 28.
Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.
Solution:

Let the line containing the point P (3, 2)
passes through x-axis at A (x, 0) and y-axis at B (0, y)
OA = OB given
∴ x = y

Question 29.
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Solution:
Three vertices of a parallelogram ABCD taken in order are
A (3, 6), B (5, 10) and C (3, 2)
(i) We need to find the co-ordinates of D
We know that the diagonals of a parallelogram bisect each other
Let (x, y) be the co-ordinates of D

Question 30.
A and B arc two points on the x-axis and y-axis respectively. P (2, – 3) is the mid point of AB. Find the

(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB. (2010)
Solution:
Points A and B are on x-axis and y-axis respectively
Let co-ordinates of A be (X, O) and of B be (O, Y)
P (2, -3) is the midpoint of AB

Question 31.
Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2 , y = – 2 and y = 6.

Solution:
The equations of sides of a rectangle whose equations are
x1 = -1, x2 = 2, y1 = -2, y2 = 6.
These lines form a rectangle when they intersect at A, B, C, D respectively
Co-ordinates of A, B, C and D will be
(-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.
AC and BD are its diagonals
(i) Slope of the diagonal AC

Question 32.
Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7
Solution:
5x + 7y = 3 …(i)
2x – 3 y = 7 …(ii)
Multiply (i) by 3 and (ii) by 7,
15x + 21y = 9
14x – 21y = 49

Question 33.
Point A (3, – 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4, 3).
(i) Write down the co-ordinates of A’ and B.
(ii) Find the slope of the line A’B, hence find its inclination.
Solution:
A’ is the image of A (3, -2) on reflection in the x-axis.
∴ Co-ordinates of A’ will be (3, 2)
Again B’ (- 4, 3) in the image of A’, when reflected in the y-axis
∴ Co-ordinates of B will be (4, 3)
(ii) Slope of the line joining, the points A’ (3, 2) and B (4, 3)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

More Exercises

Question 1.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B
Solution:
Base BC of an equilateral ∆ABC lies on y-axis
co-ordinates of point C are (0, – 3),
origin (0, 0) is the mid-point of BC.

Question 2.
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y – axis.
(ii) the image of B’ of B under reflection in the lineAA’.
(iii) the length of A’B’.
Solution:

(i) Co-ordinates of A’, the image of A (4, 3)
reflected in y-axis will be ( – 4, 3).
(ii) Co-ordinates of B’ the image of B (0, 1)
reflected in the line AA’ will be (0, 5).
(iii) Length A’B’

Question 3.
Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.
Solution:
Let R be the point whose co-ordinates are (x, y)
which divides PQ in the ratio of 3:1.

Question 4.
Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Co-ordinates of A (3, 1) and B ( – 2, 5)
P lies on AB such that

Question 5.
P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Solution:
Given
AP = PQ = QB

Question 6.
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).
Solution:
Let A (2, -2), B (8, -2) and centre of the circle be
O (α + 2, α – 5)

Question 7.
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Given
(3, 5) is the mid-point of A (2, p) and B (q, 4)

Question 8.
The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.
Solution:
Let AB be the diameter where co-ordinates of
A are (3, 0) and of B are (-5, 6).
Co-ordinates of its origin O will be

Question 9.
In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?
Solution:
Let the point (-4, 6) divides the line segment joining the points
A (-6, 10) and B (3, -8), in the ratio m : n

Question 10.
Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.
Solution:
Let P (-3, p) divides AB in the ratio of m1 : m2 coordinates of
A (-5, -4) and B (-2, 3)

Question 11.
In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Solution:
Let the point P which is on the x-axis, divides the line segment
joining the points A (4, 2) and B (3, -5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)

Question 12.
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.
Solution:
Let co-ordinates of A be (-4, 3) and of B (6, 3) and of P be (2, y)
Let the ratio in which the P divides AB be m1 : m2

Question 13.
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.
Solution:
Points are given A (2, -2), B (3, 7)
and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2
at P and let co-ordinates of

Question 14.
The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
A’ is the reflection of A(2, -3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A” is the reflection of A’ (2, 3)
∴Co-ordinates OA” will be (6, 3)
(ii) Join AA” which intersects x-axis at P whose
co-ordinate are (4, 0)
Let P divide AA” in the ratio in m1 : m2

Hence P(4, 0) divides AA” in the ratio 1 : 1

Question 15.
ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.
Solution:
Let the co-ordinates of C be (x, y) and other three vertices
of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴O is mid-points of BD
∴ Co-ordinates of O will be

Question 16.
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Solution:
ABCD is a || gm , m which co-ordinates of A are (2, -3) and B (-1, -1)
Its diagonals AC and BD bisect each other at M (1, -4)
∴ M is the midpoint of AC and BD
Let co-ordinates of C be (x1, y1) and of D be (x2, y2)
when M is the midpoint of AC then

Question 17.
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.

Solution:
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (3, 1) divides it in the ratio of 2 : 3.

Question 18.
Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.

Solution:
(i) Let co-ordinates of Q be (x’, y’) and of R (x”, y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3

Question 19.
If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Solution:
A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of ∆ABC
D, E and F are the midpoints of sides BC, CA and AB respectively
and G is the centroid of the ∆ABC

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula MCQS

More Exercises

Choose the correct answer from the given four options (1 to 12) :

Question 1.
The points A (9, 0), B (9, 6), C ( – 9, 6) and D ( – 9, 0) are the vertices of a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
Solution:
A (9, 0), B (9, 6), C (-9, 6), D (-9, 0)
AB² = (x2 – x1)² + (y2 – y1

Question 2.
The mid-point of the line segment joining the points A ( – 2, 8) and B ( – 6, – 4) is
(a) ( – 4, – 6)
(b) (2, 6)
(c) ( – 4, 2)
(d) (4, 2)
Solution:
Mid-point of the line segment joining the points A (-2, 8), B (-6, -4)

Question 3.
If $$P\left( \frac { a }{ 3 } ,4 \right)$$ segment joining the points Q ( – 6, 5) and R ( – 2, 3), then the value of a is
(a) – 4
(b) – 6
(c) 12
(d) – 12
Solution:
$$P\left( \frac { a }{ 3 } ,4 \right)$$ is mid-point of the line segment
joining the points Q (-6, 5) and R (-2, 3)

Question 4.
If the end points of a diameter of a circle are A ( – 2, 3) and B (4, – 5), then the coordinates of its centre are
(a) (2, – 2)
(b) (1, – 1)
(c) ( – 1, 1)
(d) ( – 2, 2)
Solution:
End points of a diameter of a circle are (-2, 3) and B (4,-5)
then co-ordinates of the centre of the circle
= $$\left( \frac { -2+4 }{ 2 } ,\frac { 3-5 }{ 2 } \right) or\left( \frac { 2 }{ 2 } ,\frac { -2 }{ 2 } \right)$$
= (1, -1) (b)

Question 5.
If one end of a diameter of a circle is (2, 3) and the centre is ( – 2, 5), then the other end is
(a) ( – 6, 7)
(b) (6, – 7)
(c) (0, 8)
(d) (0, 4)
Solution:
One end of a diameter of a circle is (2, 3) and centre is (-2, 5)
Let (x, y) be the other end of the diameter

Question 6.
If the mid-point of the line segment joining the points P (a, b – 2) and Q ( – 2, 4) is R (2, – 3), then the values of a and b are
(a) a = 4, b = – 5
(b) a = 6, b = 8
(c) a = 6, b = – 8
(d) a = – 6, b = 8
Solution:
the mid-point of the line segment joining the
points P (a, b – 2) and Q (-2, 4) is R (2, -3)

Question 7.
The point which lies on the perpendicular bisector of the line segment joining the points A ( – 2, – 5) and B (2, 5) is
(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) ( – 2, 0)
Solution:
the line segment joining the points A (-2, -5) and B (2, -5), has mid-point
= $$\left( \frac { -2+2 }{ 2 } ,\frac { -5+5 }{ 2 } \right)$$ = (0, 0)
(0, 0) lies on the perpendicular bisector of AB. (a)

Question 8.
The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are
(a) (x, y)
(b) (y, x)
(c) $$\left( \frac { x }{ 2 } ,\frac { y }{ 2 } \right)$$
(d) $$\left( \frac { y }{ 2 } ,\frac { x }{ 2 } \right)$$

Solution:
In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)
The point which is equidistant from O, A and B is the mid-point of AB.
∴ Coordinates are $$\left( \frac { 0+2x }{ 2 } ,\frac { 2y+0 }{ 2 } \right)$$ or (x, y) (a)

Question 9.
The fourth vertex D of a parallelogram ABCD whose three vertices are A ( – 2, 3), B (6, 7) and C (8, 3) is
(a) (0, 1)
(b) (0, – 1)
(c) ( – 1, 0)
(d) (1, 0)
Solution:
ABCD is a ||gm whose vertices A (-2, 3), B (6, 7) and C (8, 3).
The fourth vertex D will be the point on which diagonals AC and BD
bisect each other at O.

Question 10.
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, – 5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, – 5) and (2, 0)
(b) (0, 10) and ( – 4, 0)
(c) (0, 4) and ( – 10, 0)
(d) (0, – 10) and (4, 0)
Solution:
A line intersects y-axis at P and x-axis a Q.
R (2, -5) is the mid-point

Question 11.
The points which divides the line segment joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally lies in the
Solution:
A point divides line segment joining the points
A (7, -6) and B (3, 4) in the ratio 1 : 2 internally.

Let (x, y) divides it in the ratio 1 : 2

Question 12.
The centroid of the triangle whose vertices are (3, – 7), ( – 8, 6) and (5, 10) is
(a) (0, 9)
(b) (0, 3)
(c) (1, 3)
(d) (3, 3)
Solution:
Centroid of the triangle whose Vertices are (3, -7), (-8, 6) and (5, 10) is
$$\left( \frac { 3-8+5 }{ 3 } ,\frac { -7+6+10 }{ 3 } \right) or\left( 0,\frac { 9 }{ 3 } \right)$$
or (0, 3) (b)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test.

More Exercises

Question 1.
The point P (4, – 7) on reflection in x-axis is mapped onto P’. Then P’ on reflection in the y-axis is mapped onto P”. Find the co-ordinates of P’ and P”. Write down a single transformation that maps P onto P”.
Solution:
P’ is the image of P (4, -7) reflected in x-axis
∴ Co-ordinates of P’ are (4, 7)
Again P” is the image of P’ reflected in y-axis
∴ Co-ordinates of P” are (-4, 7)
∴ Single transformation that maps P and P” is in the origin.

Question 2.
The point P (a, b) is first reflected in the origin and then reflected in the y-axis to P’. If P’ has co-ordinates (3, – 4), evaluate a, b
Solution:
The co-ordinates of image of P(a, b) reflected in origin are (-a, -b).
Again the co-ordinates of P’, image of the above point (-a, -b)
reflected in the y-axis are (a, -b).
But co-ordinates of P’ are (3, -4)
∴a = 3 and -b = -4
b = 4 Hence a = 3, b = 4.

Question 3.
A point P (a, b) becomes ( – 2, c) after reflection in the x-axis, and P becomes (d, 5) after reflection in the origin. Find the values of a, b, c and d.
Solution:
If the image of P (a, b) after reflected in the x-axis be (a, -b) but it Is given (-2, c).
a = -2, c = -b
If P is reflected in the origin, then its co-ordinates will be (-a, -b), but it is given (d, 5)
∴ -b = 5 ⇒ b = -5
d = -a = -(-2) = 2, c = -b = -(-5) = 5
Hence a = -2, b = -5, c = 5, d = 2

Question 4.
A (4, – 1), B (0, 7) and C ( – 2, 5) are the vertices of a triangle. ∆ ABC is reflected in the y-axis and then reflected in the origin. Find the co-ordinates of the final images of the vertices.
Solution:
A (4, -1), B (0, 7) and C (-2, 5) are the vertices of ∆ABC.
After reflecting in y-axis, the co-ordinates of points will be
A’ (-4, -1), B’ (0, 7), C’ (2, 5). Again reflecting in origin,
the co-ordinates of the images of the vertices will be
A” (4, 1), B” (0, -7), C” (-2, -5)

Question 5.
The points A (4, – 11), B (5, 3), C (2, 15), and D (1, 1) are the vertices of a parallelogram. If the parallelogram is reflected in the y-axis and then in the origin, find the co-ordinates of the final images. Check whether it remains a parallelogram. Write down a single transformation that brings the above change.
Solution:
The points A (4, -11), B (5, 3), C (2, 15) and D (1, 1) are the vertices of a parallelogram.
After reflecting in/-axis, the images of these points will be
A’ ( -4, 11), B’ (-5, 3), C (-2, 15) and D’ (-1, 1).
Again reflecting these points in origin, the image of these points will be
A” (4, -11), B” (5, -3), C” (2, -15), D” (0, -1)
Yes, the reflection of a single transformation is in the x-axis.

Question 6.
Use a graph paper for this question (take 2 cm = 1 unit on both x and y axes).
(i) Plot the following points:
A (0, 4), B (2, 3), C (1, 1) and D (2, 0).
(ii) Reflect points B, C, D on 7-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
(iii) Join points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation of line of symmetry of the figure formed. (2017)
Solution:
(i) On graph A (0, 4), B (2, 3), C (1, 1) and D (2, 0)
(ii) B’ = (-2, 3), C’ = (-1, 1), D’ = (-2, 0)

The equation of the line of symmetry is x = 0

Question 7.
The triangle OAB is reflected in the origin O to triangle OA’B’. A’ and B’ have coordinates ( – 3, – 4) and (0, – 5) respectively.
(i) Find the co-ordinates of A and B.
(ii) Draw a diagram to represent the given information.
(iii) What kind of figure is the quadrilateral ABA’B’?
(iv) Find the coordinates of A”, the reflection of A in the origin followed by reflection in the y-axis.
(v) Find the co-ordinates of B”, the reflection of B in the x-axis followed by reflection in the origin.
Solution:
∆ OAB is reflected in the origin O to ∆ OA’B’,
Co-ordinates of A’ = (-3, -4), B’ (0, -5).
.’. Co-ordinates of A will be (3, 4) and of B will be (0, 5).
(ii) The diagram representing the given information has been drawn here.
(iii) The figure in the diagram is a rectangle.
(iv) The co-ordinates of B’, the reflection of B is the x-axis is (0, -5)
and co-ordinates of B”, the reflection in origin of the point (0, -5) will be (0, 5).
(v) The co-ordinates of the points, the reflection of A in the origin are (-3, -4)
and coordinates of A”, the reflected in y-axis of the point (-3, – 4) are (3, -4)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection MCQS

More Exercises

Choose the correct answer from the given four options (1 to 7) :

Question 1.
The reflection of the point P ( – 2, 3) in the x-axis is
(a) (2, 3)
(b) (2, – 3)
(c) ( – 2, – 3)
(d) ( – 2, 3)
Solution:
Reflection of the point P (-2, 3) in x-axis is (-2, -3) (c)

Question 2.
The reflection of the point P ( – 2, 3) in the y- axis is
(a) (2, 3)
(b) (2, – 3)
(c) ( – 2, – 3)
(d) (0, 3)
Solution:
The reflection of the point P (-2, 3) under reflection in y-axis (2, 3) (a)

Question 3.
If the image of the point P under reflection in the x-axis is ( – 3, 2), then the coordinates of the point P are
(a) (3, 2)
(b) ( – 3, – 2)
(c) (3, – 2)
(d) ( – 3, 0)
Solution:
The image of the point P under reflection in the x-axis is (-3, 2),
then the co-ordinates of the point P will be (-3, -2) (b)

Question 4.
The reflection of the point P (1, – 2) in the line y = – 1 is
(a) ( – 3, – 2)
(b) (1, – 4)
(c) (1 , 4)
(d) (1, 0)
Solution:
The reflection of the point P (1, -2) in the line y = -1 is (1, 0) (d)

Question 5.
The reflection of the point A (4, -1) in the line x = 2 is
(a) (0, – 1)
(b) (8, – 1)
(c) (0, 1)
(d) none of these
Solution:
The reflection of A (4, -1) in the line x = 2 will be A’ (0, -1) (a)

Question 6.
The reflection of the point ( – 3, 0) in the origin is the point
(a) (0, – 3)
(b) (0, 3)
(c) (3, 0)
(d) none of these
Solution:
Reflection of the point (-3, 0) in origin will be (3, 0) (c)

Question 7.
Which of the following points is invariant with respect to the line y = – 2 ?
(a) (3, 2)
(b) (3, – 2)
(c) (2, 3)
(d) ( – 2, 3)
Solution:
The variant points are (3, -2) (b)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Ex 10

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Ex 10

More Exercises

Question 1.
Find the co-ordinates of the images of the following points under reflection in the x- axis:
(i) (2, -5)
(ii) $$-\frac { 3 }{ 2 } ,-\frac { 1 }{ 2 }$$
(iii) ( – 7, 0)
Solution:
Co-ordinates of the images of the points
under reflection in the x-axis will be
(i) Image of (2, -5) will be (2, 5)
(ii) Image of $$-\frac { 3 }{ 2 } ,-\frac { 1 }{ 2 }$$ will be $$-\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }$$
(iii) Image of ( -7, 0) will be (-7, 0)

Question 2.
Find the co-ordinates of the images of the following points under reflection in the y-axis:
(i) (2, – 5)
(ii) $$-\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }$$
(iii) (0, – 7)
Solution:
Co-ordinates of the image of the points under reflection in the y-axis
(i) Image of (2, -5) will be ( -2, -5)
(ii) Image of $$-\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }$$ will be $$\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }$$
(iii) Image of (0, -7) will be (0, -7)

Question 3.
Find the co-ordinates of the images of the following points under reflection in the origin:
(i) (2, – 5)
(ii) $$\frac { -3 }{ 2 } ,\frac { -1 }{ 2 }$$
(iii) (0, 0)
Solution:
Co-ordinates of the image of the points under reflection in the y-axis
(i) Image of (2, -5) will be (-2, 5)
(ii) Image of $$\frac { -3 }{ 2 } ,\frac { -1 }{ 2 }$$ will be $$\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }$$
(iii) Image of (0, 0) will be (0, 0)

Question 4.
The image of a point P under reflection in the x-axis is (5, – 2). Write down the co-ordinates of P.
Solution:
As the image of a point (5, -2) under x – axis is P
∴ Co-ordinates of P will be (5, 2)

Question 5.
A point P is reflected in the x-axis. Co-ordinates of its image are (8, – 6).
(i) Find the co-ordinates of P.
(ii) Find the co-ordinates of the image of P under reflection in the y-axis.
Solution:
The co-ordinates of image of P which is reflected in x-axis are (8, – 6), then
(0 Co-ordinates of P will be (8, 6)
(ii) Co-ordinates of image of P under reflection in the y-axis will be ( – 8, 6)

Question 6.
A point P is reflected in the origin. Co-ordinates of its image are (2, – 5). Find
(i) the co-ordinates of P.
(ii) the co-ordinates of the image of P in the x-axis.
Solution:
The co-ordinates of image of a point P which is reflected in origin are (2, – 5), then
(i) Co-ordinates of P will be ( – 2, 5)
(ii) Co-ordinates of the image of P in the x- axis will be ( – 2, – 5)

Question 7.
(i) The point P (2, 3) is reflected in the line x = 4 to the point P’. Find the co-ordinates of the point P’.
(ii) Find the image of the point P (1, – 2) in the line x = – 1.
Solution:
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit
(b) Plot point P (2, 3) on it.
(c) Draw a line x = 4 which is parallel to y-axis.
(d) From P, draw a perpendicular on x = 4, which intersects x = 4 at Q.
(e) Produce PQ to P’, such that QP’ = QP.
∴ P’ is the reflection of P in the line x = 4
Co-ordinates of P’ are (6, 3)

(ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot the point P (1, -2) on it.
(c) Draw a line x = -1 which is parallel toy-axis.
(d) From P, draw a perpendicular on the line x = -1, which meets it at Q.
(e) Produce PQ to P’ such that PQ = QP’
P’ is the image or reflection of P in the line x = -1
Co-ordinates of P’ are (-3, -2)
Co-ordinates of P’ are (-3, -2)

Question 8.
(i) The point P (2, 4) on reflection in the line y = 1 is mapped onto P’ Find the co-ordinates of P’.
(ii) Find the image of the point P ( – 3, – 5) in the line y = – 2.
Solution:
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (2, 4) on it.
(c) Draw a line y = 1, which is parallel to x-axis.
(d) From P, draw a perpendicular on y = 1 meeting it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
P’ is the reflection of P whose co-ordinates are (2, -2)

(ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (-3, -5) on it.
(c) Draw a line y = -2 which is parallel to the x-axis.
(d) From P, draw a perpendicular on y = -2 which meets it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
Then P’ is the image of P, whose co-ordinates are (-3, 1).

Question 9.
The point P ( – 4, – 5) on reflection in y-axis is mapped on P’. The point P’ on reflection in the origin is mapped on P”. Find the co-ordinates of P’ and P”. Write down a single transformation that maps P onto P”.
Solution:
P’ is the image of point P (-4, -5) in y-axis
∴Co-ordinates of P’ will be (4, -5)
Again P” is the image of P’ under reflection in origin will be (-4, 5).
The single transformation that maps P onto P” is the x-axis

Question 10.
Write down the co-ordinates of the image of the point (3, – 2) when :
(i) reflected in the x-axis
(ii) reflected in the y-axis
(iii) reflected in the x-axis followed by reflection in the y-axis
(iv) reflected in the origin. (2000)
Solution:
Co-ordinates of the given points are (3, -2).
(i) Co-ordinates of the image reflected in x- axis will be (3, 2)
(ii) Co-ordinates of the image reflected in y- axis will be (-3, -2)
(iii) Co-ordinates of the point reflected in x- axis followed by reflection in the y-axis will be (-3, 2)
(iv) Co-ordinates of the point reflected in the origin will be (-3, 2)

Question 11.
Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by a reflection in the line x = 1.
Solution:
(i) Draw axis XOX’ and YOY’ taking 1 cm = 1 unit.
(ii) Plot a point P (3, 1).
(iii) Draw a line x = 1, which is parallel to y-axis.
(iv) From P, draw a perpendicular on x-axis meeting it at Q.
(v) Produce PQ to P’ such that QP’ = PQ, then
P’ is the image of P is x-axis. Then co-ordinates of P’ will be (3, -1)
(vi) From P’, draw a perpendicular on x = 1 meeting it at R.
(vii) Produce P’R to P” such that RP” = P’R
∴P” is the image of P’ in the line x = 1
Co-ordinates of P” are (-1, -1)

Question 12.
If P’ ( – 4, – 3) is the image of a point P under reflection in the origin, find
(i) the co-ordinates of P.
(ii) the co-ordinates of the image of P under reflection in the line y = – 2.
Solution:
(i) Reflection of P is P’ (-4, -3) in the origin
∴ Co-ordinates of P will be (4, 3)
Draw a line y = -2, which is parallel to x-axis
(ii) From P, draw a perpendicular on y = -2 meetings it at Q
Produce PQ to P” such that QP” = PQ
∴P” will the image of P in the line y = -2
∴Co-ordinates of P” will be (4, -7)

Question 13.
A Point P (a, b) is reflected in the x-axis to P’ (2, – 3), write down the values of a and b. P” is the image of P, when reflected in the y-axis. Write down the co-ordinates of P”. Find the co-ordinates of P”’, when P is reflected in the line parallel to y-axis such that x = 4. (1998)
Solution:
P’ (2, -3) is the reflection of P (a, b) in the x-axis
∴Co-ordinates of P’ will be P’ (a, – b) but P’ is (2, -3)
Comparing a = 2, b = 3

∴Co-ordinates of P will be (2, 3)
P” is the image of P when reflected in y-axis
∴Co-ordinate of P” will be ( – 2, 3)
Draw a line x = 4, which is parallel to y-axis
and P'” is the image of P when it is reflected in the line x = 4,
then P'” is its reflection Co-ordinates of P”‘ will be (6, 3).

Question 14.
(i) Point P (a, b) is reflected in the x-axis to P’ (5, – 2). Write down the values of a and b.
(ii) P” is the image of P when reflected in the y-axis. Write down the co-ordinates of P”.
(iii) Name a single transformation that maps P’ to P”. (1997)
Solution:
(i) Image of P (a, b) reflected in the x-axis to P’ (5, -2)
∴ a = 5 and b = 2
(ii) P” is the image of P when reflected in the y-axis
∴ its co-ordinates will be (-5, -2).
(iii) The single transformation that maps P’ to P” is the origin.

Question 15.
Points A and B have co-ordinates (2, 5) and (0, 3). Find
(i) the image A’ of A under reflection in the x-axis.
(ii) the image B’ of B under reflection in the line AA’.
Solution:
Co-ordinates of A are (2, 5) and of B are (0, 3)

(i) Co-ordinates of A’, the image of A reflected in the x-axis will be (2, -5)
(ii) Co-ordinates of B’, the image of B under reflection in the line AA’ will be (4, 3).

Question 16.
Plot the points A (2, – 3), B ( – 1, 2) and C (0, – 2) on the graph paper. Draw the triangle formed by reflecting these points in the x-axis. Are the two triangles congruent?

Solution:
The points A (2, -3), B (-1, 2) and C(0, -2) has been plotted on the graph paper as shown and are joined to form a triangle ABC. The co-ordinates of the images of A, B and C reflected in x-axis will be A’ (2, 3), B’ (-1, -2), C’ (0, 2) respectively and are joined to from another ∆ A’B’C’
Yes, these two triangles are congruent.

Question 17.
The points (6, 2), (3, – 1) and ( – 2, 4) are the vertices of a right angled triangle. Check whether it remains a right angled triangle after reflection in the y-axis.
Solution:
Let A (6, 2), B (3, -1) and C (-2, 4) be the points of a right-angled triangle
then the co-ordinates of the images of A, B, C reflected in y-axis be
A’ (-6, 2), B’ (-3, -1) and C’ (2, 4).

By joining these points, we find that ∆A’B’C’ is also a right angled triangle.

Question 18.
The triangle ABC where A (1, 2), B (4, 8), C (6, 8) is reflected in the x-axis to triangle A’ B’ C’. The triangle A’ B’ C’ is then reflected in.the origin to triangle A”B”C” Write down the co-ordinates of A”, B”, C”. Write down a single transformation that maps ABC onto A” B” C”.
Solution:
The co-ordinates of ∆ ABC are A (1, 2) B (4, 8), C (6, 8)
which are reflected in x- axis as A’, B’ and C’.
∴ The co-ordinates of A’ (1, -2), B (4, -8) and C (6, -8).
A’, B’ and C’ are again reflected in origins to form an ∆A”B”C”.
∴ The co-ordinates of A” will be (-1, 2), B” (-4, 8) and C” (-6, 8)
The single transformation that maps ABC onto A” B” C” is y-axis.

Question 19.
The image of a point P on reflection in a line l is point P’. Describe the location of the line l.
Solution:
The line will be the right bisector of the line segment joining P and P’.

Question 20.
Given two points P and Q, and that (1) the image of P on reflection in y-axis is the point Q and (2) the mid point of PQ is invariant on reflection in x-axis. Locate
(i) the x-axis
(ii) the y-axis and
(iii) the origin.
Solution:
Q is the image of P on reflection in y-axis
and mid point of PQ is invariant on reflection in x-axis

(i) x-axis will be the line joining the points P and Q.
(ii) The line perpendicular bisector of line segment PQ is the y-axis.
(iii) The origin will be the mid point of line segment PQ.

Question 21.
The point ( – 3, 0) on reflection in a line is mapped as (3, 0) and the point (2, – 3) on reflection in the same line is mapped as ( – 2, – 3).
(i) Name the mirror line.
(ii) Write the co-ordinates of the image of ( – 3, – 4) in the mirror line.
Solution:
The point (-3,0) is the image of point (3, 0)
and point (2, -3) is image of point (-2, -3) reflected on the same line.
(i) It is clear that the mirror line will be y-axis.
(ii) The co-ordinates of the image of the point (-3, -4)
reflected in the same line i.e. y-axis will be (3, -4).

Question 22.
A ( – 2, 4) and B ( – 4, 2) are reflected in the y-axis. If A’ and B’ are images of A and B respectively, find

(i) the co-ordinates of A’ and B’.
(ii) Assign special name to quad. AA’B’B.
(iii) State whether AB’ = BA’.
Solution:
A (-2, 4) and B (-4, 2) are reflected in y- axis as A’ and B’.

(i) Co-ordinates of A’ are (2, 4) and of B are (4, 2).
(ii) The quadrilateral AA’ B’ B is an isosceles trapezium.
(iii) yes, AB’ = BA’

Question 23.
Use graph paper for this question.
(i) The point P (2, – 4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(ii) Point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(iii) Name the figure PQR.
(iv) Find the area of figure PQR. (2007)
Solution:
(i) Since the point Q is the reflection of the point P (2, -4) in the line x = 0,
the co-ordinates of Q are (2, 4).
(ii) Since R is the reflection of Q (2, 4) about the line y = 0,
the co-ordinates of R are ( – 2, 4).
(iii) Figure PQR is the right angled triangle PQR.

(iv) Area of ∆ PQR = $$\\ \frac { 1 }{ 2 }$$ x QR x PQ
= $$\\ \frac { 1 }{ 2 }$$ x 4 x 8
= 16 sq. units.

Question 24.
Use graph paper for this question. The point P (5, 3) was reflected in the origin to get the image P’.
(i) Write down the co-ordinates of P’.
(ii) If M is the foot of perpendicular from P to the x-axis, find the co-ordinates of M.
(iii) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.
(iv) Name the figure PMP’N.
(v) Find the area of the figure PMP’N. (2001)
Solution:
P’ is the image of point P (5, 3) reflected in the origin.

(i) Co-ordinates of P’ will be (-5, -3).
(ii) M is the foot of the perpendicular from P to the x-axis.
Co-ordinates of M will be (5, 0)
(iii) N is the foot of the perpendicular from P’ to x-axis.
Co-ordinates of N will be (-5, 0).
(iv) By joining the points, the figure PMP’N is a parallelogram.
(v) Area of the parallelogram = 2 x area of ∆ MPN
= 2 x $$\\ \frac { 1 }{ 2 }$$ x MN x PM = MN x PM
= 10 x 3 = 30 sq. units. Ans.

Question 25.
Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter.
Solution:
(i) A (6, 4), B (0, 4)

Perimeter = Sum of all sides = 6 + 10 + 6 + 10 = 32 units

Question 26.
Use graph paper to answer this question
(i) Plot the points A (4, 6) and B (1, 2).
(ii) If A’ is the image of A when reflected in x-axis, write the co-ordinates of A’.
(iii) If B’ is the image of B when B is reflected in the line AA’, write the co-ordinates of B’.
(iv) Give the geometrical name for the figure ABA’B’. (2009)
Solution:
(i) Plotting the points A (4, 6) and B (1, 2) on the given graph.
(ii) A’ = (4, -6)
(iii) B’ = (7, 2)
(iv) In the quadrilateral ABA’B’, we have AB = AB’ and A’B = A’B’
Hence, ABA’B’ is a kite.

Question 27.
The points A (2, 3), B (4, 5) and C (7, 2) are the vertices’s of ∆ABC. (2006)
(i) Write down the co-ordinates of A1, B1, C1 if ∆ A1B1C1 is the image of ∆ ABC when reflected in the origin.
(ii) Write down the co-ordinates of A2, B2, C2 if ∆ A2B2C2 is the image of ∆ ABC when reflected in the x-axis.
(iii) Assign the special name to the quadrilateral BCC2B2 and find its area.

Solution:
Points A (2, 3), B (4, 5) and C (7, 2) are the vertices’s of ∆ ABC.
A1, B1 and C1 are the images of A, B and C reflected in the origin.
(i) Co-ordinates of A1 = (-2, -3) of B1 (-4, -5) and of C1 (-7, -2).
(ii) Co-ordinates of A2, B2 and C2 the images of A, B and C
when reflected in x-axis are A2 (2, – 3), B2 (4, – 5), C2 (7, – 2)
(iii) The quadrilateral formed by joining the points,
BCC2B2 is an isosceles trapezium and its area

Question 28.
The point P (3, 4) is reflected to P’ in the x-axis and O’ is the image of O (origin) in the line PP’. Find :
(i) the co-ordinates of P’ and O’,
(ii) the length of segments PP’ and OO’.
(iii) the perimeter of the quadrilateral POP’O’.
Solution:
P’ is the image of P (3, 4) reflected in x- axis
and O’ is the image of O the origin in the line P’P.
(i) Co-ordinates of P’ are (3, -4)
and co-ordinates of O’ reflected in PP’ are (6, 0)
(ii) Length of PP’ = 8 units and OO’ = 6 units
(iii) Perimeter of POP’O’ is

Question 29.
Use a graph paper for this question. (Take 10 small divisions = 1 unit on both axes). P and Q have co-ordinates (0, 5) and ( – 2, 4).
(i) P is invariant when reflected in an axis. Name the axis.
(ii) Find the image of Q on reflection in the axis found in (i).
(iii) (0, k) on reflection in the origin is invariant. Write the value of k.
(iv) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis. (2005)
Solution:

(i) Two points P (0, 5) and Q (-2, 4) are given As the abscissa of P is 0.
It is invariant when is reflected in y-axis.
(ii) Let Q’ be the image of Q on reflection in y-axis.
Co-ordinate of Q’ will be (2, 4)
(iii) (0, k) on reflection in the origin is invariant.
co-ordinates of image will be (0, 0). k = 0
(iv) The reflection of Q in the origin is the point Q”
and its co-ordinates will be (2, – 4)
and reflection of Q” (2, – 4) in x-axis is (2, 4) which is the point Q’

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Ex 10 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Chapter Test

More Exercises

Question 1.
Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.
Solution:
First 4 term of A.P. whose first term (a) = -5
and common difference (d) = -3
= -5, -8, -11, -14

Question 2.
Verify that each of the following lists of numbers is an A.P., and the write its next three terms :
(i) $$0,\frac { 1 }{ 4 } ,\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,…$$
(ii) $$5,\frac { 14 }{ 3 } ,\frac { 13 }{ 3 } ,4,…$$
Solution:
(i) $$0,\frac { 1 }{ 4 } ,\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,…$$
Here a = 0, d = $$\\ \frac { 1 }{ 4 }$$

Question 3.
The nth term of an A.P. is 6n + 2. Find the common difference.
Solution:
Tn of an A.P. = 6n + 2 .
T1 = 6 x 1 + 2 = 6 + 2 = 8
T2 = 6 x 2 + 2 = 12 + 2 = 14
T3 = 6 x 3 + 2 = 18 + 2 = 20
d = 14 – 8 = 6

Question 4.
Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16th term and the nth.
Solution:
9, 12, 15, 18, …
Here, a = 9, d = 12 – 9 = 3
or 15 – 12 = 3
or 18 – 15 = 3
Yes, it form an A.P.

Question 5.
Find the 6th term from the end of the A.P. 17, 14, 11, …, – 40.
Solution:
6th term from the end of
A.P. = 17, 14, 11, …… 40
Here, a = 17, d = -3, l = -40
l = a + (n – 1 )d

Question 6.
If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.
Solution:
In an A.P.
a8 = 31, a15 = a11 + 16
Let a be the first term and d be a common difference, then

Question 7.
The 17th term of anA.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the wth term.
Solution:
In an A.P.
a17 = 2 x a8 + 5
a11 = 43, find an
Let a be the first term and d be the common

Question 8.
The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.
Solution:
In an A.P.
a19 = 3 x a6 and a9 = 19
Let a be the first term and d be the common
difference, then

Question 9.
If the 3rd and the 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Solution:
In an A.P.
a3 = 4, a9 = -8, which term of A.P. will be zero
Let a be the first term and d be a common difference, then

Question 10.
Which term of the list of numbers 5, 2, – 1, – 4, … is – 55?
Solution:
A.P. is 5, 2, -1, – 4, …
Which term of A.P. is -55
Let it be nth term
Here, a = 5, d = 2 – 5 = -3

Question 11.
The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.
Solution:
In an A.P.
24th term = 2 x 10th term
To show that 72nd term = 4 x 15th term
Let a be the first term and d be a common difference, then

Question 12.
Which term of the list of numbers $$20,19\frac { 1 }{ 4 } ,18\frac { 1 }{ 2 } ,17\frac { 3 }{ 4 } ,..$$ is the first negative term?
Solution:
In A.P., which is the first negative term
$$20,19\frac { 1 }{ 4 } ,18\frac { 1 }{ 2 } ,17\frac { 3 }{ 4 } ,..$$

Question 13.
If the pth term of an A.P. is q and the qth term is p, show that its nth term is (p + q – n)
Solution:
In an A.P.
pth term = q
qth term = p
Show that (p + q – n) is nth term
Let a be the first term and d be the common

Question 14.
How many three digit numbers are divisible by 9?
Solution:
3-digit numbers which are divisible by 9 are 108, 117, 126, 135, …, 999
Here, a = 108, d = 9 and l = 999

Question 15.
The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.
Solution:
Sum of three numbers of an A.P. = -3
and their product = 8
Let the numbers be
a – d, a, a + d, then
a – d + a + a + d = -3

Question 16.
The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.
Solution:
Angles of a quadrilateral are in A.P.
Greatest angle is double of the smallest
Let the smallest angle of the quadrilateral is
a + 3d…..(i)

Question 17.
The nth term of an A.P. cannot be n² + n + 1. Justify your answer.
Solution:
nth term of an A.P. can’t be n² + n + 1
Giving some different values to n such as 1, 2, 3, 4, …
we find then

Question 18.
Find the sum of first 20 terms of an A.P. whose nth term is 15 – 4n.
Solution:
Giving some different values such as 1 to 20
We get,

Question 19.
Find the sum :
$$18+15\frac { 1 }{ 2 } +13+…+\left( -49\frac { 1 }{ 2 } \right)$$
Solution:
Find the sum
$$18+15\frac { 1 }{ 2 } +13+…+\left( -49\frac { 1 }{ 2 } \right)$$

Question 20.
(i) How many terms of the A.P. – 6,$$– \frac { 11 }{ 2 }$$ – 5,… make the sum – 25?
(ii) Solve the equation 2 + 5 + 8 + … + x = 155.
Solution:
(i) Sum = -25
A.P. = -6, $$– \frac { 11 }{ 2 }$$ -5,…

Question 21.
If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.
Solution:
3rd term of an A.P. = 5
Ratio in 6th term and 10th term = 7 : 13
Find S20
Let a be the first term and d be the common difference

Question 22.
In an A.P., the first term is 2 and the last term is 29. If the sum of the terms is 155, then find the common difference of the A.P.
Solution:
In an A.P.
First term (a) = 2
Last term (l) = 29
Sum of terms = 155

Question 23.
The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.
Solution:
Sum of first 14 terms = 1505
First term (a) = 10
Find 25th term

Question 24.
Find the number of terms of the A.P. – 12, – 9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
Solution:
A.P. -12, -9, -6,…, 21
If 1 is added to each term, find the sum of there terms

Question 25.
The sum of first n term of an A.P. is 3n² + 4n. Find the 25th term of this A.P.
Solution:
S= 3n² + 4n
Sn – 1 = 3(n – 1)² + 4(n – 1)

Question 26.
In an A.P., the sum of first 10 terms is – 150 and the sum of next 10 terms is – 550. Find the A.P.
Solution:
In an A.P.
Sum of first 10 terms = -150
Sum of next 10 terms = -550, A.P. = ?
Sum of first 10 terms = -150

Question 27.
The sum of first m terms of an A.P. is 4m² – m. If its nth term is 107, find the value of n. Also find the 21 st term of this A.P.
Solution:
Sm = 4m² – m
Sn = 4n² – n

Question 28.
If the sum of first p, q and r terms of an A.P. are a, b and c respectively, prove that
$$\frac { a }{ p } (q-r)+\frac { b }{ q } (r-p)+\frac { c }{ r } (p-q)=0$$
Solution:
Let the first term of A.P. be A and common difference be d.
Sum of the first p terms is

Question 29.
A sum of Rs 700 is to be used to give 7 cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
What is the importance of an academic prise in students life? (Value Based)
Solution:
Total sum = Rs 700
Cash prizes to 7 students = 7 prize
Each prize is Rs 20 less than its preceding prize
d = -20, d = -20, n = 7

Question 30.
Find the geometric progression whose 4th term is 54 and 7th term is 1458.
Solution:
In a G.P.
a4 = 54
a7 = 1458
Let a be the first term and r be the common difference

Question 31.
The fourth term of a G.P. is the square of its second term and the first term is – 3. Find its 7th term.
Solution:
In G.P.
a4 = (a2)², a1 = -3
Let a be the first term and r be the common ratio

Question 32.
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.
Solution:
In a G.P.
a= x, a10 = y, a16 = z
Show that x, y, z are in G.P.
Let a be the first term and r be the common ratio, then

Question 33.
The original cost of a machine is Rs 10000. If the annual depreciation is 10%, after how many years will it be valued at Rs 6561 ?
Solution:
Original cost of machine = Rs 10000
Since, machine depreciates at the rate of 10%
on reducing the balance,
Value of machine after one year

Question 34.
How many terms of the G.P. $$3,\frac { 3 }{ 2 } ,\frac { 3 }{ 4 }$$,are needed to give the sum $$\\ \frac { 3069 }{ 512 }$$ ?
Solution:
G.P. $$3,\frac { 3 }{ 2 } ,\frac { 3 }{ 4 }$$
Sn = $$\\ \frac { 3069 }{ 512 }$$

Question 35.
Find the sum of first n terms of the series : 3 + 33 + 333 + …
Solution:
Series is
3 + 33 + 333 + … n terms
= 3[1 + 11 + 111 +…n terms]

Question 36.
Find the sum of the series 7 + 7.7 + 7.77 + 7.777 + … to 50 terms.
Solution:
The given sequence is 7, 7.7, 7.77, 7.777,…
Required sum = S50
= 7 + 7.7 + 7.77 + … 50 terms
= 7(1 + 1.1 + 1. 11 + … 50 terms)

Question 37.
The inventor of chessboard was a very clever man. He asked the king, h reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third, and so on, doubling the number of the grains for each subsequent square. How many grains would have to be given?
Solution:
In a chessboard, there are 8 x 8 = 64 squares
If a man put 1 grain in first square,
2 grains in second square,
4 grains in third square
and goes on upto the last square, i.e. 64th square
Therefore, 1 + 2 + 4 + 8 + 16 + … 64 terms
Here, a = 1, r = 2 and n = 64

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions MCQS

More Exercises

Choose the correct answer from the given four options (1 to 33) :

Question 1.
The list of numbers – 10, – 6, – 2, 2, … is
(a) an A.P. with d = – 16
(b) an A.P with d = 4
(c) an A.P with d = – 4
(d) not an A.P
Solution:
-10, -6, -2, 2, … is
an A.P. with d = – 6 – (-10)
= -6 + 10 = 4 (b)

Question 2.
The 10th term of the A.P. 5, 8, 11, 14, … is
(a) 32
(b) 35
(c) 38
(d) 185
Solution:
10th term of A.P. 5, 8, 11, 14, …
{∵ a = 5, d = 3}
a + (n – 1)d = 5 + (10 – 1) x 3
= 5 + 9 x 3
= 5 + 27
= 32 (a)

Question 3.
The 30th term of the A.P. 10, 7, 4, … is
(a) 87
(b) 77
(c) – 77
(d) – 87
Solution:
30th term of A.P. 10, 7, 4, … is
30th term = a + (n – 1)d

Question 4.
The 11th term of the A.P. – 3, $$– \frac { 1 }{ 2 }$$, 2, … is
(a) 28
(b) 22
(c) – 38
(d) – 48
Solution:
Given
-3, $$– \frac { 1 }{ 2 }$$, 2, …

Question 5.
The 4th term from the end of the A.P. – 11, – 8, – 5, …, 49 is
(a) 37
(b) 40
(c) 43
(d) 58
Solution:
4th term from the end of the A.P. -11, -8, -5, …, 49 is
Here, a = -11, d = -8 – (-11) = -8 + 11 = 3 and l = 49 .

Question 6.
The 15th term from the last of the A.P. 7, 10, 13, …,130 is
(a) 49
(b) 85
(c) 88
(d) 110
Solution:
15th term from the end of A.P. 7, 10, 13,…, 130
Here, a = 7, d = 10 – 7 = 3, l = 130
15th term from the end = l – (n – 1)d
= 130 – (15 – 1) x 3
= 130 – 42
= 88 (c)

Question 7.
If the common difference of an A.P. is 5, then a18 – a13 is
(a) 5
(b) 20
(c) 25
(d) 30
Solution:
Common difference of an A.P. (d) = 5
a18 – a13 = a + 17d – a – 12d = 5d = 5 x 5 = 25 (c)

Question 8.
In an A.P., if a18 – a14 = 32 then the common difference is
(a) 8
(b) – 8
(c) – 4
(d) 4
Solution:
If a18 – a14 = 32, then d = ?
(a + 17d) – a – 13d = 32
⇒ a + 17d – a – 13d = 32
⇒ 4d = 32
⇒ $$d \frac { 32 }{ 4 }$$ = 8 (a)

Question 9.
In an A.P., if d = – 4, n = 7, an = 4, then a is
(a) 6
(d) 7
(c) 20
(d) 28
Solution:
In an A.P., d = -4, x = 7, an = 4 then a = ?
an = a(n – 1)d = 4
a7 = a + (7 – 1)d = 4
⇒ a + 6d = 4
⇒ a + 6 x (-4) = 4
a – 24 = 4
⇒ a = 4 + 24 = 28 (d)

Question 10.
In an A.P., if a = 3.5, d = 0, n = 101, then an will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Solution:
In an A.P.
a = 3.5, d= 0, n = 101, then an = ?
an = a101 = a + (101 – 1)d
= 3.5 + 100d
= 3.5 + 100 x 0
= 3.5 + 0
= 3.5 (b)

Question 11.
In an A.P., if a = – 7.2, d = 3.6, an = 7.2, then n is
(a) 1
(b) 3
(c) 4
(d) 5
Solution:
In an A.P.
a = – 7.2, d = 3.6, an = 7.2, n = ?
an = 7.2
a + (n – 1)d = 7.2
– 7.2 + (n – 1) 3.6 = 7.2
(n – 1) x 3.6 = 7.2 + 7.2 = 14.4
(n – 1) = $$\\ \frac { 14.4 }{ 3.6 }$$ = 4
n = 4 + 1 = 5 (d)

Question 12.
Which term of the A.P. 21, 42, 63, 84,… is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Solution:
Which term of an A.P. 21, 42, 63, 84, … is 210
Let 210 be nth term, then
Here, a = 21, d = 42 – 21 =21
210 = a + (n – 1)d
210 = 21 + (n – 1) x 21
⇒ 210 – 21 = 21(n – 1)
⇒ $$\\ \frac { 189 }{ 21 }$$ = n – 1
⇒ 9 = n – 1
⇒ n = 9 + 1 = 10
.’. It is 10th term. (b)

Question 13.
If the last term of the A.P. 5, 3, 1, – 1,… is – 41, then the A.P. consists of
(a) 46 terms
(b) 25 terms
(c) 24 terms
(d) 23 terms
Solution:
Last term of an A.P. 5, 3, 1, -1, … is -41
Then A.P. will consist of ……. terms
Here, a = 5, d = 3 – 5 = – 2 and n =?
l = -41
l = -41
l = -41 = a + (n – 1 )d
-41 = 5 + (n – 1) (-2)
-41 – 5 = (n – 1) (-2)
⇒ $$\\ \frac { -46 }{ -2 }$$ = n – 1
⇒ n – 1 = 23
⇒ n = 23 + 1 = 24
A.P. consists of 24 terms. (c)

Question 14.
If k – 1, k + 1 and 2k + 3 are in A.P., then the value of k is
(a) – 2
(b) 0
(c) 2
(d) 4
Solution:
k – 1, k + 1 and 2k + 3 are in A.P.
2(k+ 1) = (k – 1) + (2k + 3)
⇒ 2k + 2 = k – 1 + 2k + 3
⇒ 2k + 2 – 3k + 2
⇒ 3k – 2k = 2 – 2
⇒ k = 0 (b)

Question 15.
The 21st term of an A.P. whose first two terms are – 3 and 4 is
(a) 17
(b) 137
(c) 143
(d) – 143
Solution:
First two terms of an A.P. are – 3 and 4
a = -3, d = 4 – (-3) = 4 + 3 = 7
21st term = a + 20d
= -3 + 20(7)
= -3 + 140
= 137 (b)

Question 16.
If the 2nd term of an A.P. is 13 and the 5th term is 25, then its 7th term is
(a) 30
(b) 33
(c) 37
(d) 38
Solution:
In an A.P.
2nd term = 13 ⇒ a + d = 13 …(i)
5th term = 25 ⇒ a + 4d = 25 …(ii)
Subtracting (i) and (ii),
3d = 12
⇒ d = $$\\ \frac { 1 }{ 3 }$$
Substitute the value of d in eq. (i), we get
a = 13 – 4 = 9
7th term = a + 6d = 9 + 6 x 4 = 9 + 24 = 33 (b)

Question 17.
If the first term of an A.P. is – 5 and the common difference is 2, then the sum of its first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Solution:
First term (a) of an A.P. = -5
Common difference (d) is 2
Sum of first 6 terms = $$\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right]$$
= $$\frac { 6 }{ 2 } \left[ 2\times \left( -5 \right) +\left( 6-1 \right) \times 2 \right]$$
= 3[-10 + 5 x 2]
= 3 x [-10 – 10]
= 3 x 0 = 0 (a)

Question 18.
The sum of 25 terms of the A.P.$$-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 }$$ is
(a) 0
(b) $$– \frac { 2 }{ 3 }$$
(c) $$– \frac { 50 }{ 3 }$$
(d) – 50
Solution:
Sum of 25 terms of an A.P.
$$-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 }$$ is

Question 19.
In an A.P., if a = 1, an = 20 and Sn = 399, then n is
(a) 19
(b) 21
(c) 38
(d) 42
Solution:
In an A.P., a = 1, an = 20, Sn = 399, n is ?
an = a + (n – 1 )d = 20
1 +(n – 1)d = 20
(n – 1)d = 20 – 1 = 19 …(i)

Question 20.
In an A.P., if a = – 5, l = 21. and Sn = 200, then n is equal to
(a) 50
(b) 40
(c) 32
(d) 25
Solution:
In an A.P.
a = -5, l = 21, Sn = 200, n = ?
l = a + (n – 1)d = -5 + (n – 1 )d
21 = -5 + (n – 1)d

Question 21.
In an A.P., if a = 3 and S8 = 192, then d is
(a) 8
(b) 7
(c) 6
(d) 4
Solution:
In an A.P.
a = 3, S8 =192, d = ?

Question 22.
The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
Solution:
First 5 multiples of 3 :
3, 6, 9, 12, 15
Here, a = 3, d = 6 – 3 = 3

Question 23.
The number of two digit numbers which are divisible by 3 is
(a) 33
(b) 31
(c) 30
(d) 29
Solution:
Two digit number which are divisible by 3 is 12, 15, 18, 21, … 99
Here, a = 12, d = 3, l = 99
l = an = a + (n – 1)d
⇒ 12 + (n – 1) x 3 = 99
⇒ (n – 1)3 = 99 – 12 = 87
⇒ n – 1 = $$\\ \frac { 87 }{ 3 }$$ = 29
⇒ n = 29 + 1 = 30 (c)

Question 24.
The number of multiples of 4 that lie between 10 and 250 is
(a) 62
(b) 60
(c) 59
(d) 55
Solution:
Multiples of 4 lying between 10 and 250 12, 16, 20, 24, …, 248
Here, a = 12, d = 16 – 12 = 4, l = 248

Question 25.
The sum of first 10 even whole numbers is
(a) 110
(b) 90
(c) 55
(d) 45
Solution:
Sum of first 10 even whole numbers
Even numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18
Here, a = 0, d = 2, n = 10

Question 26.
The list of number $$\\ \frac { 1 }{ 9 }$$ , $$\\ \frac { 1 }{ 3 }$$, 1, – 3,… is a
(a) GP. with r = – 3
(b) G.P. with r = $$– \frac { 1 }{ 3 }$$
(c) GP. with r = 3
(d) not a G.P.
Solution:
The given list of numbers
$$\\ \frac { 1 }{ 9 }$$ , $$\\ \frac { 1 }{ 3 }$$, 1, – 3,…

Question 27.
The 11th of the G.P. $$\\ \frac { 1 }{ 8 }$$ , $$– \frac { 1 }{ 4 }$$ , 2, – 1, … is
(a) 64
(b) – 64
(c) 128
(d) – 128
Solution:
11th of the G.P.
$$\\ \frac { 1 }{ 8 }$$ , $$– \frac { 1 }{ 4 }$$ , 2, -1, … is

Question 28.
The 5th term from the end of the G.P. 2, 6, 18, …, 13122 is
(a) 162
(b) 486
(c) 54
(d) 1458
Solution:
5th term from the end of the G.P. 2, 6, 18, …, 13122 is
Here, a = 2, r = $$\\ \frac { 6 }{ 2 }$$ = 3, l = 13122

Question 29.
If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is
(a) – 1
(b) – 4
(c) 1
(d) 4
Solution:
k, 2(k + 1), 3(k + 1) are in G.P.
[2(k + 1)]² = k x 3(k + 1)
⇒ 4(k + 1)² = 3k(k + 1)
⇒ 4 (k + 1) = 3 k
(Dividing by k + 1 if k + 1 ≠ 0) 4
⇒ 4k + 4 = 3k
⇒ 4k – 3k = -4
⇒ k = -4 (b)

Question 30.
Which term of the G.P. 18, – 12, 8, … is $$\\ \frac { 512 }{ 729 }$$ ?
(a) 12th
(b) 11th
(c) 10th
(d) 9th
Solution:
Which term of the G.P.
18, -12, 8,… $$\\ \frac { 512 }{ 729 }$$
Let it be nth term

Question 31.
The sum of the first 8 terms of the series 1 + √3 + 3 + … is
Solution:
Sum of first 8 terms of 1 + √3 + 3 + … is
Here a = 1, $$r=\frac { \sqrt { 3 } }{ 1 } =\sqrt { 3 }$$ , n = 8

Question 32.
The sum of first 6 terms of the G.P. 1, $$– \frac { 2 }{ 3 }$$ ,$$\\ \frac { 4 }{ 9 }$$ ,… is
(a) $$– \frac { 133 }{ 243 }$$
(b) $$\\ \frac { 133 }{ 243 }$$
(c) $$\\ \frac { 793 }{ 1215 }$$
(d) none of these
Solution:
Sum of first 6 terms of G.P.
1, $$– \frac { 2 }{ 3 }$$ ,$$\\ \frac { 4 }{ 9 }$$ ,…

Question 33.
If the sum of the GP., 1,4, 16, … is 341, then the number of terms in the GP. is
(a) 10
(b) 8
(c) 6
(d) 5
Solution:
The sum of G.P. 1, 4, 16, … is 341
Let n be the number of terms,

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.