## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B.

Other Exercises

Question 1.
Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3), (-1, 7)
Solution:
Let P (x, y) be the mid-point in each case

Question 2.
Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Solution:
Co-ordinates of A (3, 5), B (x, y) and mid-point M (2, 3)

Question 3.
A (5, 3), B (-1, 1) and C (7, -3) are the vertices of ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = $$\frac { 1 }{ 2 }$$ BC.
Solution:

Question 4.
Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10),
(ii) B; if A = (3, -1) and M (-1, 3).
Solution:
M is the mid-point of AB.
(i) Let A = (x, y), M = (1, 7) and B = (-5, 10)

Question 5.
P (-3, 2) is the mid-point of line segment AB as shown in the figure. Find the co-ordinates of points A and B.

Solution:

Point A is on y-axis
its abscissa is zero and point B is on x-axis
its ordinate is zero.
Now, let co-ordinates of A are (0, y) and ofB are (x, 0) and P (-3, 2) is the mid-point

Question 6.
In the given figure, P (4, 2) is the mid point of line segment AB. Find the co-ordinates of A and B.

Solution:

Points A and B are on x-axis and y-axis respectively
Ordinate of A is zero and abscissa of B is zero.
Let co-ordinates of A be (x, 0) and B (0, y)
and P (4, 2) is the mid-point

Question 7.
(-5, 2), (3, -6) and (7, 4) arc the vertices of a triangle. Find the length of its median through the vertex (3, -6) and (7, 4).
Solution:

Let A (-5, 2), B (3, -6) and C (7, 4) are the vertices of a ABC
Let L,M and N are the mid-points of sides BC, CA and AB respectively of ABC.
L is the mid-point of BC.
Co-ordinates of L will be

Question 8.
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution:

Question 9.
One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).
Solution:

Question 10.
A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Solution:
Co-ordinates of A = (2, 5), B = (1, 0), C = (-4, 3) and D = ( 3, 8)

Let the mid-point of AC is P (x1, y1) Co-ordinates of mid-point of AC will be

Co-ordinates of mid-points AC and BD are the same..

Question 11.
P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of its diagonals. Find the co-ordinates of R and S.
Solution:
In the parallelogram PQRS and qo-ordinates of P are (4, 2) and of Q are (-1, 5).
The diagonals of || gm AC and BD intersect each other at O (-3, 2)

Question 12.
A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates of vertex C.
Solution:

Vertices of a parallelogram ABCD are A (-1, 0), B (1, 3) and D(3, 5)
Let co-ordinates of C be (x, y)
Let the diagonals AC and BD bisect each other at O. Then O is the mid-point of AC as well as of BD.
Co-ordinates of O, the mid-point of BD will be

Question 13.
The points (2, -1), (-1, 4) and (-2, 2) are the mid-points of the sides of a triangle. Find its vertices.
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of a ABC respectively.

Co-ordinates of A are (-5, 7), of B are (1, -3) and of C are (3, 1)

Question 14.
Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Solution:

Question 15.
Points P (a, -4), Q (-2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of ‘a’ and ‘b’:
Solution:

Question 16.
Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C = (-1, 4).
Solution:

Question 17.
The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Solution:

Question 18.
A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A.

Other Exercises

Question 1.
Calculate the co-ordinates-of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1 : 2
(ii) A (-4, 6) and B (3, -5) in the ratio 3 : 2
Solution:
(i) Let co-ordinates of P be (x,y)

Question 2.
In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis ?
Solution:
Let the point P (x, 0) divides in the ratio of m1 : m2 line joining the points A (2, -3) and B (5, 6)

Question 3.
In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis ?
Solution:
Let the point P (0, y) divides the line joining the points A (2, -4) and (-3, 6) in the ratio of m1 : m2

Question 4.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1)? Also, find the value of ‘a’.
Solution:
Let the point P (1, a) divides the line joining the points (-1, 4) and (4, -1) in the ratio of m1 : m2

Question 5.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of ‘a’.
Solution:
Let the point P (a, 6) divides the line joining the points A (-4, 3), B (2, 8) in the ratio of m1 : m2

Question 6.
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Solution:

Question 7.
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Let, the points (0, y) be the point of intersection which divides the line joining the points A (-4, 7) and B (3, 0)

Question 8.
Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Solution:
Points A, B, C and D divide the line segment joining the points (5, -10) and origin (0, 0) in five equal parts
Let co-ordinates of A be (x, y) which divides PO in the ratio of 1 : 4

Question 9.
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that = $$\frac { PA }{ PB }$$ = $$\frac { 1 }{ 5 }$$, find the co-ordinates of P.
Solution:
Let the co-ordinates of P be (x, y) which divides the line joining the points A (-3,-10) and B (-2,6) in the ratio of AP : PB i.e. (5 – 1) : 1 or 4 : 1

Question 10.
P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Solution:

Question 11.
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Solution:
Let the point P (2, y) divides the line joining the points A (-3, -1) and B (5, 7) in the ratio of m1 : m2

Question 12.
Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Solution:
Let the point P (x, 2) divides the line joining the points A (6, 5) and B (4, -3) in the ratio of m1 : m2

Question 13.
The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.

Solution:
From the figure, the line AB intersects x-axis at A and y-axis at B.
Let the co-ordinates of A (x, 0) and B (0, y) and P (5, -4) divides it in the ratio of 2 : 5

Question 14.
Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Solution:

Question 15.
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Solution:

Let the points A (-5, 8) and B (10, -4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
AP = PQ = QB
AP : PB = 1 : 2 and AQ : QB = 2 : 1

Question 16.
Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Solution:

Let A and B are the points of trisection of the line segment joining the points P (2, 1) and Q (5, -8), then
PA = AB = BQ.
PA : AQ = 1 : 2 and PB : BQ = 2 : 1

Question 17.
If A = (-4, 3) and B = (8, -6)
(i) find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis ?
Solution:

Question 18.
The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Solution:

Question 19.
A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that:
AP : PB = AQ : QC = 1 : 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = $$\frac { 1 }{ 3 }$$ BC.
Solution:

Question 20.
A (-3, 4), B ( 3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2 : 3.
Solution:

Question 21.
ThelinesegmentjoiningA(2, 3)andB(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the pointK. [1990, 2006]
Solution:
Let the line segment Intersect the x-axis at the point P
Co-ordinates of P are (x, 0)
Let P divide the line segment in the ratio K : 1 then

Question 22.
The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Solution:
Points A (4, -7), B (-6, -2) are joined which intersects y-axis at K. abscissa of K will be 0

Let the coordinates of K be (0, y) and K divides AB line segment in the ratio m1 : m2

Question 23.
The line joining P (-4, 5) and Q (3, 2), intersects they axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) The ratio PR: RQ.
(ii) The co-ordinates of R.
(iii) The areas of the quadrilateral PMNQ. [2004]
Solution:
(i) Let divides the line joining the points P (-4, 5) and Q (3, 2) in the ratio k : 1

Question 24.
In the given figure, line APB meets the x- axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.

Solution:
Let the co-ordinates of A be (x1, 0) (as it lies on x-axis)
and co-ordinates of B be (0, y2)
and co-ordinates of P are (-4, 2)
AP : PB = 1 : 2 i.e. m1 = 1, m2 = 2
Now, P divides AB in the ratio m1 : m2 or 1 : 2

Question 25.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Ex 12B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12B.

Other Exercises

Question 1.
Attempt this question on graph paper.
(a) Plot A (3, 2) and B (5, 4) on graph paper. Take 2cm = 1 unit on both the axes.
(b) Reflect A and B in the x-axis to A’ and B’ respectively. Plot these points also on the same graph paper.
(c) Write down:
(i) The geometrical name of the figure ABB’A’;
(ii) The measure of angle ABB’ ;
(iii) The image A” of A, when A is reflected in the origin.
(d) The single transformation that maps A’ to A”. [1995]
Solution:
From the graph, we can say that
(iii) (a) an isosceles trapezium
(b) 45°
(c) ( -3, -2)
(d) reflection in y-axis

Question 2.
Points (3,0) and (-1,0) are invariant points under reflection in the line L1; points (0, -3) and (0, 1) are invariant points on reflection in line L2
(a) Name or write equations for the lines L1 and L2 .
(b) Write down the images of points P (3, 4) and Q ( -5, -2) on reflection in L. Name the images as P’ and Q’ respectively.
(c) Write down the images of P and Q on re-flection in L,. Name the images as P” and Q” respectively.
(d) State or describe a single transformation that maps P’ onto P”. [1996]
Solution:

(a) Points (3, 0) and (-1, 0) are invariant points under reflection L .
Here the ordinates of both points are 0 It is x-axis or y = 0.
Again points (0, -3) and (0, 1) are invariant- points on reflection in line L2
Here the abscissas are 0 in both points.
It is y-axis or x = 0.
(b) The co-ordinates of the images of points P (3, 4) and Q (-5, -2) on reflection in L, will be P (3, -4) and Q'(-5, 2)
(c) The co-ordinate of images of points P (3, 4) and Q (-5, -2) on reflection in L2 will be P’ (-3, 4) and Q’ (5, -2)
(d) Reflection is in origin because the co-ordinates of P’ and P” have opposite signs.

Question 3.
(a) Point P (a, b) is reflected in the x-axis to P’ (5, -2), write down the values of a and b.
(b) P” is the image of P when reflected in y- axis, write down the co-ordinates of P”.
(c) Name a single transformation that maps P’ to P”. (1997):
Solution:
(a) P (a, b) is reflected in x-axis to P’ (5, -2). the co-ordinates of P will be (5, 2)
a = 5, b = 2
(b) P” is the image of P when reflected is y-axis.
Co-ordinates of P” will be ( -5, 2)
(c) Reflection is in origin.

Question 4.
The point (-2, 0) on reflection in a line is mapped to (2, 0) and the point (5, -6) on reflection in the same line is mapped to (-5, -6).
(a) State the name of the mirror line and write its equation.
(b) State the co-ordinates of the image of (-8, -5) in the mirror line.
Solution:
Point (-2, 0) is mapped to (2, 0) to a certain line and point (5, -6) is also mapped to (-5, -6) to the same line.
(a) We see that sign of x-coordinate is changed.
The required line is y-axis whose equation will be x = 0.
(b) The co-ordinates of the image of the point (-8, -5) in the same mirror line will be (8, -5).

Question 5.
The points P (4, 1) and Q (-2, 4) are reflected in line y = 3, Find the co-ordinates of P’, the image of P and Q’, the image of Q.
Solution:
Co-ordinates of P and Q are (4, 1) and ( -2, 4) respectively.

The co-ordinates of image of P which is P’ are (4, 5) reflection in the line y = 3
and co-ordinates of image of Q when is Q’ are ( -2, 2) reflection is the line y = 3 as shown in the graph.

Question 6.
A point P (-2, 3) is reflected in the line x = 2 to point P’. Find the co-ordinates of P’.
Solution:
The image of P ( -2, 3) is P’ which is reflected in the line x = 2. The co-ordinates of P’ will be (6, 3) as shown in the graph.

Question 7.
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, 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:

(i) A point P (a, b) is reflected in v-axis to P’ (2, -3)
Co-ordinates of P will be (2, 3) a = 2, b = 3.
(ii) P” is the image of P, when reflected in the y- axis
Co-ordinates of P” will be ( -2, 3)
(iii) The image of P is P'” reflected in a line parallel to y-axis i.e. x = 4.
Co-ordinates of P”‘ will be (6, 3)

Question 8.
Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image:
(a) A’ of A under reflection in the x-axis.
(b) B’ of B under reflection in the line AA’.
(c) A” of A under reflection in the y-axis.
(d) B” of B under reflection in the line AA”.
Solution:
Co-ordinates of A are (3, 4) and co-ordinates of B are (0, 2)
(a) Co-ordinates, of A’ of A under reflection in the v-axis are (3, -4)
(b) Co-ordinates of B’ of B under reflection in the line AA’ are (6, 2)
(c) Co-ordinates of A” of A under reflection in y-axis are (-3, 4)
(d) Co-ordinates of B” of B under reflection in the line A A” are (0, 8).

Question 9.
(a) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.
(b) A’ is the image of A when reflected in the v-axis. Write down the co-ordinates of A’ and plot it on the graph paper.
(c) B’ is the image of B when reflected in the y-axis, followed by reflection in the origin.
Write down the co-ordinates of B’ and plot it on the graph paper.
(d) Write down the geometrical name of the figure AA’ BB’.
(e) Name two invariant points under reflection in the x-axis. [1999]
Solution:
(a) The points A (3, 5) and B(-2, -4) have been plotted on the graph.
(b) A’ is the image of A when reflected in the x-axis

Co-ordinates of A’ will be (3, -5)
(c) B’ is the image of B when reflected in y- axis, followed by reflection in the origin.
Co-ordinates will be (-2, 4)
Images A’ and B’ have also been plotted.
(d) Join AA’, A’B, BB’ and B’A the quadrilateral so joined is an isosceles trapezium.
(e) The points C and D whose co-ordinates ! are (3,0) and (-2, 0) are invariant points under reflection in the x-axis.

Question 10.
The point P (5,3) was reflected in the origin to get the image P’.
(a) Write down the co-ordinates of P’.
(b) If M is the foot of the perpendicular from P to the x-axis, find the co-ordinates of M.
(c) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.
(d) Name the figure PM P’ N.
(e) Find the area of the figure PM P’ N.
Solution:
(a) Points P(5, 3) is reflected to P’ in the origin.
Co-ordinates of P’ will be (-5, -3).
(b) PM ⊥ on x-axis.
Co-ordinates of P’ will be (5, 0)
(c) P’N X on x-axis
Coordinates of N will be (-5, 0).
(d) PM, MP’ P’N and N. P are joined which form a parallelogram PMP’N.
(e) Area of || gm PMP’N = 2 (Area of ∆PMN)
= 2 ($$\frac { 1 }{ 2 }$$ x 10 x 3) Sq. cm
= 30 Sq. cm

Question 11.
The point P (3, 4) is reflected to P’ in the x-axis; and O’ is the image of O (the origin) when reflected in the line PP’. Write :
(i) the co-ordinates of P’ and O’,
(ii) the length of the segments PP’ and OO’,
(iii) the perimeter of the quadrilateral POP’O’.
(iv) the geometrical name of the figure POP’O’.
Solution:

(i) Point P (3, 4) is reflected to P’ in x-axis and then its co-ordinates will be (3, -4) again O’ is the image of O (0, 0) reflected in PP’, then co-ordinates of O’ will be (6, 0)
(ii) Length of PP’ = 4 + 4 = 8 units and OO’ = 3 + 3 = 6 units
(iii) Diagonals of OPO’P’ bisect each other at right angles at M.
OPO’P’ is a rhombus.
(iv) The perimeter of quadrilateral OPO’P’ = 5 x 4 = 20 units
(v) The quadrilateral OPO’P’ is a rhombus as rhombus shown on the graph paper.

Question 12.
A (1, 1), B (5, 1), C (4, 2) and D (2, 2) are vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C, and D are reflected in the origin on to A’, B’, C’ and D’ respectively. Locate A’, B’, C’ and D’ on the graph sheet and write their co-ordinates. Are D, A, A’ and D’ collinear? (2004)
Solution:
On the given graph, plot the points A(1, 1), B(5, 1). C(4, 2) and D(2, 2) Join AB, BC, CD and DA.

The quadrilateral is of trapezium in shape.
Now points A, B, C and D are reflected in the origin on to A’, B’, C and D’ respectively.
Co-ordinates of A’ are (-1, -1), of B’ are (-5, -1), of C are (-4, -2) and of D’ are (-2, -2).
A’ B, B’C’, CD’ and D’A’ are joined.
Yes, we see that DA A’D’ are collinear.

Question 13.
P and Q have co-ordinates (0, 5) and (-2, 4).
(a) P is invariant when reflected in an axis, Name the axis.
(b) Find the image of Q on reflection in the axis found in fa)
(c) (0, k) on reflection in the origin is invariant. Write the value of k.
(d) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis. (2005)
Solution:
Co-ordinates of the given points P and Q, are (0, 5) and (-2, 4) and have been plotted on the graph.

(a) P is invariant in y-axis
(b) Image of Q(-2, 4) is Q’ in years and its Co-ordinates will be (2, 4)
(c) (0, k) on reflections in we origin is also invariant
Co-ordinates will be (0, 0) k = 0
(d) Let Q” be image of Q reflection in his origin.
Again Q” is reflected in x-axis and its image will be Q’ (2, 4)

Question 14.
(a) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(b) The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(c) Name the figure PQR.
(d) Find the area of figure PQR.
Solution:

The graph of the solution set is shown by the thick portion of the number line. The solid circle at -3 indicates that the number – 3 is included among the solutions whereas the open circle at 3 indicates that 3 is not included among the solutions.
(c) (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 x = 0, the co-ordinates of R are (- 2, 4).
(iii) Figure PQR is the right angled triangle PQR.
(d) Area of ∆PQR = $$\frac { 1 }{ 2 }$$ x QR x PQ = $$\frac { 1 }{ 2 }$$ x 4 x 8 = 16 sq. units.

Question 15.
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:

Question 16.
Use graph paper for this question. (Take 2 cm = 1 unit along both x and y axis.
Plot the points O (0, 0), A (-4, 4), B (-3, 0) and C (0, -3)
(i) Reflect points A and B on the y-axis and name them A’ and B’ respectively. Write down their coordinates. k
(ii) Name the figure OABCB’A’.
(iii) State the line of symmetry of this figure. (2016)
Solution:

(ii) The figure OABCB’A’ is similar to arrow headed.
(iii) The y-axis is the line of symmetry of figure OABCB’A’.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12B are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Ex 12A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12A.

Other Exercises

Question 1.
Complete the following table.

Solution:
We can complete the blanks as under
(a) Reflection in origin
(b) (4, -2)
(c) (0, 6)
(d) Reflection in origin
(e) Reflection in y-axis.

Question 2.
A point P is its own image under the reflection in a line l. Describe the position of the point P with respect to the line l.
Solution:

The image of the point is itself the point P with respect to a line l, Then the point lies in the line l.

Question 3.
State the co-ordinates of the following points under reflection in x-axis:
(i) (3, 2)
(ii) (-5, 4)
(iii) (0, 0)
Solution:
The co-ordinates of the given points under reflection in x-axis are as under :
(i) (3, -2)
(ii) (-5, -4)
(iii) (0, 0).

Question 4.
State the co-ordinates of the following points under reflection in y-axis :
(i) (6, -3)
(ii) (-1, 0)
(iii) (-8, -2)
Solution:
The co-ordinates of the points under reflection in y-axis are :
(i) (-6, -3)
(ii) (1, 0)
(iii) (8, -2)

Question 5.
State the co-ordinates of the following points under reflection in origin :
(i) (-2, -4)
(ii) (- 2, 7)
(iii) (0, 0)
Solution:
The co-ordinates of the points under reflection in origin are :
(i) (2, 4)
(ii) (2, -7)
(iii) (0, 0)

Question 6.
State the co-ordinates of the following points under reflection in the line x = 0 :
(i) ( -6, 4)
(ii) (0, 5)
(iii) (3, -4)
Solution:
The co-ordinates of the points under reflection in the line x = 0, or y-axis are :
(i) (6, 4)
(ii) (0, 5)
(iii) ( -3, -4)

Question 7.
State the co-ordinates of the following points under reflection in the line y = 0;
(i) (-3, 0)
(ii) (8, -5)
(iii) (-1, -3)
Solution:
The co-ordinates of the points under reflection in the line y = 0 or x-axis are :
(i) (-3, 0)
(ii) (8, 5)
(iii) (-1, 3)

Question 8.
A point P is reflected in the x-axis. Co-ordinates of its image are (- 4, 5).
(i) Find the co-ordinates of P.
(ii) Find the co-ordinates of the image of P under reflection in the y-axis.
Solution:
(i) The co-ordinates of a point P in the x-axis are (-4, 5).
the co-ordinates of P will be (-4, -5)
(ii) Now the co-ordinates of the image of P under reflection in the y-axis will be (4, -5).

Question 9.
A point P is reflected in the origin. Co-ordinates of its image are ( -2, 7).
(i) Find the co-ordinate of P.
(ii) Find the co-ordinates of the image of P under reflection in the x-axis.
Solution:
The co-ordinates of a point P in the origin are (-2, 7)
The co-ordinates of P will be (2, -7)
(ii) and the co-ordinate of the image of P under the reflection in the x-axis will be (2, 7)

Question 10.
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 (4, 6) evaluate a and b.
Solution:
The co-ordinates of P reflected in the origin will be P ( -a, -b).
Again the co-ordinates of image of P’ under reflection in the y-axis is rule be (a, -b).
But the co-ordinates of P’ are given as (4, 6)
a = 4, -b = 6 ⇒ b = -6

Question 11.
The point P (x, y) is first reflected in the x- axis and then reflected in the origin to P’. If P’ has co-ordinates ( -8, 5); evaluate x and y.
Solution:
The co-ordinates of P reflected to the x-axis will be (x, -y)
and its coordinate will be reflected in the origin P (-x, y).
But the co-ordinates of P’ are (-8, 5)
-x = -8 => x = 8
y = 5

Question 12.
The point A ( -3, 2) is reflected in the x- axis to the point A’. Point A’ is then reflected in the origin to point A’.
(a) Write down the co-ordinates of A’
(b) Write down a single transformation that maps A onto A’
Solution:
(a) The co-ordinates of A’, the reflected point in x-axis will be ( -3, -2).
(i) the co-ordinates of A’, the reflected points of A’ in the origin will be (3, 2).
(b) The reflection is in y-axis.

Question 13.
The point A (4, 6) is first reflected in the origin to point A’. Point A’ is then reflected in the y-axis to point A’.
(i) Write down the co-ordinates of A’.
(ii) Write down a single transformation that maps A onto A’.
Solution:
Co-ordinates of point A are (4, 6). The co-ordinate of A’, the image of A as reflected in the origin will be ( -4, -6).
(i) Again the co-ordinates of A’, the image of A’ as reflected in the y-axis will be (4, -6).
(ii) The reflection is in x-axis.

Question 14.
The triangle ABC, where A is (2, 6). B is (-3, 5) and C is (4, 7), is reflected in the y-axis to triangle A’B’C’. Triangle A’B’C’ is then reflected in the origin to triangle A”B”C”.
(i) Write down the co-ordinates of A”. B” and C”.
(ii) Write down a single transformation that maps triangle ABC onto triangle A”B”C”.
Solution:
In ∆ABC. co-ordinates of vertices.
A are (2. 6), B ( -3. 5) and C (4, 7).
Then the co-ordinates of vertices of ∆A’B’C’ when reflected in the y-axis will be A’ ( -2, 6). B’ (3, 5) and C’ ( -4, 7).
Again the co-ordinates of vertices of ∆A”B”C” when reflected in the origin will be A” (2, -6), B” ( -3, -5) and C”(4, -7).
(ii) The reflection of ∆ABC is in x-axis.

Question 15.
P and Q have co-ordinates (-2, 3) and (5, 4) respectively. Reflect P in the x-axis to P’ and Q in the y-axis to Q’. State the co-ordinates of P’ and Q’.
Solution:
Co-ordinates of P is ( -2, 3) and co-ordinates of Q (5, 4).
Then the co-ordinates of P’ when reflected in x-axis will be (-2, -3)
and the co-ordinates of Q’ when reflected in y-axis will be (-5, 4)

Question 16.
On a graph paper, plot the triangle ABC whose vertices are at the points A (3,1), B (5,0) and C (7,4). On the same diagram, draw the image of the triangle ABC under reflection in the origin 0 (0, 0).
Solution:
∆ABC whose vertices are A (3, 1), B (5, 0) and C (7, 4).
Now the image of ∆ABC is ∆A’B’C’ under reflection in the origin O (0, 0).
The co-ordinates of the vertice A’, B’ and C’ will be A’ ( -3, -1), B’ (- 5, 0) and C’ ( -7, -4) as shown on the graph.

Question 17.
Find the image of point (4, -6) under the following operations :

Solution:
Since the co-ordinates of a given point are (4, -6), then

Question 18.
Point A (4, -1) is reflected as A’ in the y- axis. Point B on reflection in the x-axis is mapped as B’ ( -2, 5). Write the co-ordinates of A’ and B.
Solution:
A’ is the reflection of point A (4, -1) in the y-axis and its co-ordinates will be (-4, -1)
Co-ordinates of B’, the image of point B are ( -2, 5) in the x-axis.
Co-ordinates of B will be ( -2, -5)

Question 19.
The point (-5, 0) on reflection in a line is mapped as (5, 0) and the point (-2, -6) on reflection in the same line is mapped as (2, -6).
(a) Name the line of reflection.
(b) Write the co-ordinates of the image of (5, -8) in the line obtained in (a).
Solution:
(a) The point (-5, 0) is reflected in a line as (5, 0)
Here, x is mapped as -x
Again point (-2, -6) is mapped in the reflection of the same line as (2, -6)
Here x is mapped as -x
The line of reflection will be y-axis
(b) Now the co-ordinates of the image of the point (5, -8) in the same line will be (-5, -8)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D.

Other Exercises

Question 1.
Find x and y, if:

Solution:

Question 2.
Find x and y, if :

Solution:

Question 3.

Solution:

Question 4.

(i) the order of the matrix X
(ii) the matrix X.
Solution:

Question 5.
Evaluate

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

(i) A (BA)
(ii) (AB) A
Solution:

Question 9.
Find x and y, if

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:
(i) Order of matrix A is 2 x 2
Order of matrix B is 2 x 1
Order of matrix X is 2 x 1

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C.

Other Exercises

Question 1.
Evaluate if possible :

If not possible, give a reason.
Solution:

(iv) It is not possible, because number of columns of the first matrix is not equal to number of rows of the second matrix.

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Find x and y, if:

Solution:

Question 5.

Solution:

Question 6.

(i) AB
(ii) BA
(iii) A²
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Comparing the elements, we get:
-2b =-2
b = 1
a = 2

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Simplify : A² + BC.
Solution:

Question 16.
Solve for x and y :

Solution:

Question 17.
In each case given below, find :
(a) the order of matrix M.
(b) the matrix M.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, What can you say about the matrix A?
Solution:
AB = BA = B
But it is possible, when A = 0 or B = 0
But B is not a zero matrix (given)
A is a zero matrix or A is an identity matrix

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.
State, with reason, whether the following are true or false. A, B and C are matrices of order 2 x 2.
(i) A + B = B + A
(ii) A – B = B – A
(iii) (B . C). A = B . (C . A)
(iv) (A + B) . C = A . C + B . C
(v) A . (B – C) = A . B – A . C
(vi) (A – B) . C = A . C – B . C
(vii) A² – B² = (A + B) (A – B)
(viii) (A – B)² = A² – 2 A . B + B²
Solution:
(i) True : Because addition of matrices is commutative.
(ii) False : Subtraction of matrices is not commutative.
(iii) True : Multiplication of matrices is associative.
(iv) True: Multiplication of matrices is distributive over addition.
(v) True : As given above in (iv)
(vi) True : As given above in (iv)
(vii) False : Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False, As given above in (vii)

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B.

Other Exercises

Question 1.
Evaluate

Solution:

Question 2.
Find x and y if :

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If I is the unit matrix of order 2 x 2; find the matrix M, such that:

Solution:

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A.

Other Exercises

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 x 2 and 2 x 3 respectively; then their sum A + B is possible,
(ii) The matrices A 2 x3, and B 2 x 3, are conformable for subtraction,
(iii) Transpose of a 2 x 1 matrix is a 2 x 1 matrix,
(iv) Transpose of a square matrix is a square matrix,
(v) A column matrix has many columns and only one row.
Solution:
(i) False, because the orders of both the matrices are not same,
(ii) True,
(iii) False, because transpose of a 2 x 1 matrix is a 1 x 2 matrix,
(iv) True,
(v) False, because it has only one column and may have many rows.

Question 2.

Solution:
Comparing the elements in order, we get that :
x = 3
y + 2 = 1 ⇒ y = 1 – 2 = -1
z – 1 = 2 ⇒ z = 2 + 1 = 3
Hence x = 3, y = -1 , z = 3

Question 3.
Solve for a, b and c; if :

Solution:
(i) Comparing the elements in order, we get
-4 = b + 4 ⇒ – b = 4 + 4 = 8 ⇒ b = – 8
a + 5 = 2 ⇒ a = 2 – 5= -3
2 = c – 1 ⇒ c = 2 + 1 = 3
Hence a = -3, b = -8, c = 3
(ii) Comparing the elements in order, we get
a = 3
a – b = – 1 ⇒ 3 – b = -1 ⇒ -b = -1 – 3 ⇒ -b = -4 ⇒ b = 4
b + c = 2 ⇒ 4 + c = 2 ⇒ c = 2 – 4 = -2
Hence a = 3, b = 4, c = – 2

Question 4.

Solution:

Question 5.

Solution:

Question 6.
Wherever possible, write each of the following as a single matrix.

Solution:

This is not possible because both the matrices are not of the same order.

Question 7.
Find, x and y from the following equations:

(ii) [-8 x] + [y -2] = [-3 2]
Solution:

Comparing, the elements of two equal martrices:
3 – x = 7 ⇒ x = -7 + 3 = -4
and y + 2 = 2 ⇒ y = 2 – 2 = 0
Hence x = – 4, y = 0
(ii) [-8 x] + [y -2] = [-3 2]
⇒ [-8 + y x – 2] = [-3 2]
Comparing the elements, we get :
-8 + y = -3 ⇒ y = -3 + 8 = 5
x – 2 = 2 ⇒ x = 2 + 2 = 4
Hence x = 4, y = 5

Question 8.

Solution:

Question 9.
Write the additive inverse of matrices A, B and C

Solution:

Question 10.
Given A = [ 2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
Solution:
A = [2 -3], B = [0 2], C = [-1 4]
(i) X + B = C – A
X + [0 2] = [-1 4] – [2 -3]
⇒ X + [0 2] = [- 1 – 2 4 – (-3)]
X + [0 2] = [-3 4 + 3] = [-3 7]
X = [-3 7] – [0 2] = [-3 – 0 7 – 2] = [-3 5]
(ii) A – X = B + C
⇒ -X = B + C – A
⇒ X = A – B – C = [2 -3] – [0 2] – [-1 4] = [2 – 0 – (-1) -3 – 2 – 4] = [ 2 + 1 – 9] = [3 – 9]

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C.

Other Exercises

Question 1.
Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
Solution:

Question 2.
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely. (2014)
Solution:
f(x) = x3 + 10x2 – 37x + 26
f(1) = (1)3 + 10(1)2 – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f(x)

Question 3.
When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x3 + 3x2 – mx + 4
and x – 2 = 0 then x = 2
f(2) = (2)3 + 3(2)2 – m(2) + 4 = 8 + 12 – 2m + 4 = 24 – 2m
Remainder = 24 – 2m
But, remainder is given m + 3
m + 3 = 24 – 2m
⇒ m + 2m = 24 – 3
⇒ 3m = 21
⇒ m = 7
Hence m = 7

Question 4.
What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor ?
Solution:
The number to be subtracted = Remainder obtained by dividing 3x3 – 8x2 + 4x – 3 by x + 2
Let f(x) = 3x3 – 8x2 + 4x – 3
and x + 2 = 0, then x = – 2
Remainder = f(-2) = 3 (-2)3 – 8 (-2)2 + 4 (-2) – 3 = -24 – 32 – 8 – 3 = -67
Hence the number to be subtracted = – 67

Question 5.
If (x + 1) and (x – 2) are factors of x3 + (a + 1) x2 – (b – 2) x – 6, find the values of a and 6. And then, factorise the given expression completely.
Solution:

Question 6.
If x – 2 is a factor of x2 – ax + b and a + b = 1, find the values of a and b.
Solution:
(x – 2) is a factor of x2 + ax + b
Let x – 2 = 0 ⇒ x = 2
Now x2 + ax + b = (2)2 + a x 2 + b = 4 + 2a + b = 2a + b + 4
x – 2 is the factor Remainder = 0 or 2a + b + 4 = 0
⇒ 2a + b = -4 …(i)
But a + b = 1 (given) …(ii)
Subtracting, we get : a = -5
Substituting the value of a in (ii)
-5 + b = 1 ⇒ b = 1 + 5 ⇒ b = 6
Hence a = -5, b = 6

Question 7.
Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Solution:

Question 8.
Find the value of ‘m’ if mx3 + 2x2 – 3 and x2 – mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx3 + 2x2 – 3
g (x) = x2 – mx + 4
Let x – 2 = 0, then x = 2
f(2) = m (2)3 + 2 (2)2 – 3 = 8m + 8 – 3 = 8m + 5
g(2) = (2)2 – mx2 + 4 = 4 – 2m + 4 = 8 – 2m
In both cases the remainder is same
8m + 5 = 8 – 2m
⇒ 8m + 2m = 8 – 5
⇒ 10m = 3
⇒ m = $$\frac { 3 }{ 10 }$$

Question 9.
The polynomial px3 + 4x2 – 3x + q is completely divisible by x2 – 1; find the values of p and q. Also, for these values of p and q, factorize the given polynomial completely.
Solution:

Question 10.
Find the number which should be added to x2 + x + 3 so that the resulting polynomial is completely divisible by (x + 3).
Solution:
Let k be added to f(x), then f(x) = x2 + x + 3 + k
Let x + 3 = 0, then x = -3
f(-3) = (-3)2 + (-3) + 3 + k = 9 – 3 + 3 + k = 9 + k
f(x) is divisible by x + 3, then remainder will be 0.
9 + k = 0 ⇒ k = -9

Question 11.
When the polynomial x3 + 2x2 – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x3 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution:
Let f(x) = x3 + 2x2 – 5ax – 1
and let x – 1 = 0, then x = 1
f(1) = (1)3 + 2(1)2 – 5a x 1 – 7 = 1 + 2 – 5a – 7 = -5a – 4
-5a – 4 = A ….(i)
Let g (x) = x3 + ax2 +12x + 16
and let x + 2 = 0, then x = -2
g (-2) = (-2)3 + a (-2)2 – 12 (-2) + 16 = -8 + 4a + 24 + 16 = 32 + 4a
32 + 4a = B ….(ii)
2A + B = 0
2 (-5a – 4) + 32 + 4a = 0
⇒ -10a – 8 + 32 + 4a = 0
⇒ -6a + 24 = 0
⇒ 6a = 24
⇒ a = 4
a = 4

Question 12.
(3x + 5) is a factor of the polynomial (a – 1) x3 + (a + 1) x2 – (2a + 1) x – 15. Find the value of ‘a’. For this value of ‘a’, factorise the given polynomial completely.
Solution:
Let f(x) = (a – 1) x3 + (a + 1) x2 – (2a + 1) x – 15

Question 13.
When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p.’
Solution:
When (x – 3) divides x3 – px2 + x + 6,
then Remainder = p(3) = (3)3 – p(3)2 + (3) + 6 = 27 – 9p + 9 = 36 – 9p
When (x – 3) divides 2x3 – x2 – (p + 3) x – 6,
then Remainder = p(3) = 2(3)3 – (3)2 – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p
⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
⇒ p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression : 2x3 + x2 – 13x + 6
Solution:
(a) By hit and trial, putting x = 2, we have
2 (8) + 4 – 26 + 6 = 0
⇒ (x – 2) is the factor of 2x3 + x2 – 13x + 6

Question 15.
Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
Let f(x) = 2x3 + 3x2 – kx + 5 By the remainder theorem,
f(2) = 7
⇒ 2(2)3 + 3(2)2 – k(2) + 5 = 7
⇒ 2(8) + 3(4) – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ 2k = 16 + 12 + 5 – 7
⇒ 2k = 26
⇒ k = 13
The value of k is 13.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B.

Other Exercises

Question 1.
Using the Factor Theorem, show that:
(i) (x – 2) is a factor of x3 – 2x2 – 9x +18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2. completely.
(iv) 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence, factorise the given expression completely.
Solution:

Question 2.
Using the Remainder Theorem, factorise each of the following completely:
(i) 3x3 + 2x2 – 19x + 6
(ii) 2x3 + x2 – 13x + 6
(iii) 3x3 + 2x2 – 23x – 30
(iv) 4x3 + 7x2 – 36x – 63
(v) x3 + x2 – 4x – 4. (2004)
Solution:

Question 3.
Using the Remainder Theorem factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation. 3x3 + 10x2 + x – 6 = 0
Solution:

Question 4.
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = o.
Solution:

Question 5.
Given that x – 2 and x +1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution:

Question 6.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Solution:

Question 7.
If x + a is a common factor of expressions f(x) = x2 + px + q and g (x) = x2 + mx + n show that a = $$\frac { n – q }{ m – p }$$
Solution:

Question 8.
The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
Solution:

Question 9.
Find the value of ‘a’ if (x – a) is a factor of x3 – ax2 + x + 2.
Solution:

Question 10.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Solution:

⇒ 9 – k = 0 ⇒ k = 9
Hence 9 should be subtracted.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A.

Other Exercises

Question 1.
Find in each case, the remainder when :
(i) x4 – 3x2 + 2x + 1 is divided by x – 1.
(ii) x3 + 3x2 – 12x + 4 is divided by x – 2.
(iii) x4 + 1 is divided by x + 1.
(iv) 4x3 – 3x2 + 2x – 4 is divided by 2x + 1.
(v) 4x3 + 4x2 – 21x + 16 is divided by 2x – 3.
(vi) 2x3 + 9x2 – x – 15 is divided by 2x + 3.
Solution:

Question 2.
Show that:
(i) x – 2 is a factor of 5x2 + 15x – 50.
(ii) 3x + 2 is a factor of 3x2 – x – 2.
(iii) x + 1 is a factor of x3 + 3x2 + 3x + 1.
Solution:

Question 3.
Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.
(i) x + 1
(ii) 2x – 1
(iii) x + 2
(iv) 3x – 2
(v) 2x – 3.
Solution:
(i) f(x) = 2×3 + 3×2 – 5x – 6

Question 4.
(i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.
Solution:

Question 5.
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression, x3 + ax2 + bx – 12.
Solution:

(x + 3) is a factor of f(x)
f(-3) = 0
9a – 3b – 39 – 0
⇒ 3a – b – 13 = 0
⇒ 3a – b = 13
Adding (i) and (ii) we get:
5a = 15 ⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = – 4
Hence a = 3, b = – 4

Question 6.
Find the value of k, if 2x + 1 is a factor of (3k + 2) x3 + (k – 1).
Solution:

Question 7.
Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Solution:

Question 8.
Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.
Solution:

Question 9.
When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constants.
Solution:
f(x) = x3 + 2x2 – kx + 4

Question 10.
Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
Solution:

Question 11.
If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b. [2005]
Solution:

Question 12.
The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
Solution:

Question 13.
What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8 ?
Solution:

Question 14.
What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it by x – 2, the remainder is 10.
Solution:

Question 15.
The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1) x – 16 leave the same remainder when divided by x – 2. Find the value of ‘a’.
Solution:

Question 16.
If (x – 2) is a factor of the expression 7x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Solution:

Question 17.
Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3. (2015)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A 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.