## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E.

**Other Exercises**

- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

**Question 1.**

**Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. ****Also, find the equation of the line through P and parallel to 3x + 5y = 7**

**Solution:**

**Question 2.**

**The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. ****Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.**

**Solution:**

Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3

Let co-ordinates of P be (x, y), then

**Question 3.**

**A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.**

**Solution:**

P lies on y-axis and let the co-ordinates of P be (0, y)

P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.

5 x 0 + 3y + 15 = 0

⇒ 3y = -15

⇒ y = -5

Co-ordinates of P are (0, -5)

Now, writing the line x – 3y + 4 = 0 is form of y = mx + c

-3y = -x – 4

⇒ 3y = x + 4

⇒ y = \(\frac { 1 }{ 3 }\) x + \(\frac { 4 }{ 3 }\)

**Question 4.**

**Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]**

**Solution:**

Writing, the line kx – 5y + 4 = 0 in form of y = mx + c

⇒ -5y = -kx – 4

⇒ 5y = kx + 4

**Question 5.**

**A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:**

**(i) the equation of the line.**

**(ii) the co-ordinates of A and B.**

**(iii) the co-ordinates of M. (2003)**

**Solution:**

(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.

Substituting, the co-ordinates in (i)

x + 0 = 3 ⇒x = 3

Co-ordinates of A are (3, 0)

Again 0 + y = 3 ⇒ y = 3

Co-ordinates of B are (0,3)

(iii) M is the mid-point of AB.

Co-ordinates of M wil be (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

**Question 6.**

**(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.**

**Solution:**

Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)

⇒ 3y – 6 = -2x – 2

⇒ 2x + 3y = 6 – 2

⇒ 2x + 3y = 4

**Question 7.**

**Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.**

**(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.**

**(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.**

**Solution:**

Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)

Let, co-ordinates of fourth vertex D be (x, y)

**Question 8.**

**A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.**

**Solution:**

Slope of line x = 3y + 2 or 3y = x – 2 ….(i)

**Question 9.**

**A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.**

**Solution:**

Let, the line intersects x-axis at A and y-axis at B.

Let, co-ordinates of A (x, o) and of (o, y)

But (3, 2) is the mid-point of AB.

**Question 10.**

**Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.**

**Solution:**

7x + 6y = 71 ….(i)

5x – 8y = -23 ….(ii)

Multiply (i) by 4 and (ii) by 3,

28x + 24y = 284

15x – 24y = -69

On adding (i) and (ii), we get:

43x = 215

x = 5

Substituting, the value of x in (i)

7 x 5 + 6y = 71

35 + 6y = 71

⇒ 6y = 71 – 35 = 36

⇒ y = 6

Point of intersection of these lines is (5, 6)

Now slope of line 4x – 2y = 1

⇒ 4x – 1 = 2y

⇒ y = 2x – \(\frac { 1 }{ 2 }\) is 2

Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is \(\frac { -1 }{ 2 }\)

Equation of the line y – y_{1} = m (x – x_{1})

⇒ y – 6 = \(\frac { -1 }{ 2 }\) (x – 5)

⇒ 2y – 12 = -x + 5

⇒ x + 2y = 5 + 12 = 17

⇒ x + 2y = 17

**Question 11.**

**Find the equation of the line which is perpendicular to the line \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 1 at the point where this line meets y-axis.**

**Solution:**

**Question 12.**

**O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:**

**(i) the equation of median of ∆OAB through vertex O.**

**(ii) the equation of altitude of ∆OAB through vertex B.**

**Solution:**

(i) Let, mid-point of AB be D.

**Question 13.**

**Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.**

**Does line 3x = y + 1 bisect the line segment joining the two given points ?**

**Solution:**

Slope of the line joining the points (-2, 3) and (4, 1)

Yes, these are perpendicular to each other

Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)

Co-ordinates of P will be

This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point

now, substituting the value of x and y. in 3x = y + 1

⇒ 3 x 1 = 2 + 1

⇒ 3 = 3. which is true.

Yes, the line 3x = y + 1 is the bisector.

**Question 14.**

**Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).**

**Solution:**

Equation of the given line is x cos 30° + y sin 30° = 2

y sin 36° = -x cos 30° + 2

**Question 15.**

**Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :**

**(i) perpendicular to the line 2x – y + 7 = 0**

**(ii) parallel to it.**

**Solution:**

Writing the given equation in the form of y = mx + c

(k – 2) x + (k + 3) y – 5 = 0 ….(i)

⇒ (k + 3) y = – (k – 2) x + 5

**Question 16.**

**The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :**

**(i) the equation of line through A and perpendicular to BC.**

**(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.**

**Solution:**

Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)

(ii) Let, the line through A meets BC in P

P is point of intersection of these two lines.

3x – 4y = 5 ……… (i)

4x + 3y = 15 …….. (ii)

On solving (i), (ii) we get

x = 3, y = 1

Co-ordinates of Pare (3, 1)

**Question 17.**

**From the given figure, find :**

**(i) the co-ordinates of A, B and C.**

**(ii) the equation of the line through A and parallel to BC. (2005)**

**Solution:**

(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)

(ii) Slope of line BC is (m)

⇒ x + 2y – 6 – 2 = 0

⇒ x + 2y – 8 = 0

⇒ x + 2y = 8

**Question 18.**

**P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)**

**Solution:**

Let (x, y) be the co-ordinates of M, the mid-point of PQ.

⇒ x + 2y – 8 = 0

⇒ x + 2y = 8

**Question 19.**

**A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.**

**Solution:**

In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.

P and Q are the mid-points of AB and AC respectively

Co-ordinates of P will be

Slopes of PQ and BC are same.

These are parallel to each other.

Quad. PBCQ is trapezium

**Question 20.**

**A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :**

**(i) the co-ordinates of A and B.**

**(ii) equation of line through P, and perpendicular to AB.**

**Solution:**

Line AB intersects x-axis at A and y-axis at B.

(i) Let co-ordinates of A be (x, 0) and of B be (0, y)

Point P (-4, -2) intersects AB in the ratio 1 : 2

**Question 21.**

**A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)**

**Solution:**

Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.

**Question 22.**

**Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)**

**Solution:**

x-intercept = 4

Co-ordinates of that point = (4, 0)

The co-ordinates of the given point (2, 3)

**Question 23.**

**The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)**

**Solution:**

Equation of line AB is y = x + 1

and equation of line CD is y = √3 x – 1

Slope of AB = 1

tanθ = 1

⇒ θ = 45°

Inclination angles of AB = 45°

Slope of CD = tanθ = √3 = tan 60°

⇒ θ = 60°

Inclination angle of CD = 60°

In ΔPQR,

Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)

⇒ 60° = 45° + θ

⇒ θ = 60° – 45° = 15°

⇒ θ = 15°

**Question 24.**

**Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)**

**Solution:**

P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3

Co-ordinates of P will be

**Question 25.**

**The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.**

**Solution:**

Let points are A (6, 4) and B (7, -5)

**Question 26.**

**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’ of (-2, p) lies on it.**

**Solution:**

Coordinates of A are (7, -3), of B = ( 1, 9)

**Question 27.**

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

**(i) Coordinates of A and B.**

**(ii) Slope of line AB.**

**(iii) equation of line AB.**

**Solution:**

As P (2, -3) is mid-point of AB.

Let coordinates of B be (0, y) and coordinates of A be (x, 0)

**Question 28.**

**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.**

**Solution:**

Given line 3x + 4y -7 = 0

⇒ 3 (y – 4) = 4 (x – 2)

⇒ 3y – 12 = 4x – 8

⇒ 4x – 3y – 8 + 12 = 0

⇒ 4x – 3y + 4 = 0

**Question 29.**

**ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :**

**(i) co-ordinates of A**

**(ii) equation of diagonal BD.**

**Solution:**

(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)

The diagonals of ||gm bisect each other

O is said point of AC and BD

Now if O is mid point of BD then its co-ordinates will be

⇒ y + 4 = 4x – 8

⇒ 4x – y -8 – 4 = 0

⇒ 4x – y – 12 = 0 or 4x – y = 12

**Question 30.**

**Given equation of line L _{1} is y = 4.**

**(i) Write the slope of line L**

_{2}if L_{2}is the bisector of angle O.**(ii) Write the co-ordinates of point P.**

**(iii) Find the equation of L**

_{2}.**Solution:**

(i) Equation of line L

_{1}is y = 4

L

_{2}is the bisector of ∠O

**Question 31.**

**Find:**

**(i) equation of AB**

**(ii) equation of CD**

**Solution:**

Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.

**Question 32.**

**Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1**

**Solution:**

x-intercept of the line = -3

and is perpendicular to the line

3x + 5y = 1

5y = 1 – 3x

**Question 33.**

**A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :**

**(i) the equation of the line.**

**(ii) the co-ordinates of points A and B.**

**(iii) the co-ordinates of point M.**

**Solution:**

A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.

M is mid-point of AB

(i) Equation of line AB will be

y – y_{1} = m (x – x_{1})

⇒ y – 4 = -1 (x + 1)

⇒ y – 4 = – x – 1

⇒ y + x = -1 + 4

⇒ x + y = 3

(ii) The line intersect x-axis at A and OA = 3 units

Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units

Co-ordinates of B are (0, 3)

(iii) M is mid-point of AB

Co-ordinates of M are (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

**Question 34.**

**In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:**

**(i) equation of line AB.**

**(ii) equation of line CD.**

**(iii) co-ordinates of points E and D.**

**Solution:**

In the given figure, AB meets y-axis at point A.

Line through C (2, 10) and D intersects line AB at P at right angle.

Equation of CD

y – 10 = 3 (x – 2)

⇒ y – 10 = 3x – 6

⇒ 3x – y + 10 – 6 = 0

⇒ 3x – y + 4 = 0

(iii) Co-ordinates of D which is on x-axis

3x – y + 4 = 0

3x – 0 + 4 = 0

⇒ 3x + 4 = 0

⇒ 3x = -4

⇒ x = \(\frac { -4 }{ 3 }\)

Co-ordinates of D are (\(\frac { -4 }{ 3 }\) , 0)

E is also on x-axis

x + 3y = 18

Substituting, y = 0, then

x + 0= 18

⇒ x = 18

Co-ordinates of E are (18, 6)

**Question 35.**

**A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.**

**Solution:**

A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.

Draw BC || x-axis

and PC || y-axis

In ∆PAD and ∆PBC

∠P = ∠P (common)

∠D = ∠C (each 90°)

∆PAD ~ ∆PBC

PB = 2PA ⇒ PA = \(\frac { 1 }{ 2 }\) PB

2y – 6 = 3x – 12

⇒ 3x – 2y – 12 + 6 = 0

⇒ 3x – 2y – 6 = 0

⇒ 3x – 2y = 6

**Question 36.**

**Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.**

**Solution:**

Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°

**Question 37.**

**Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.**

**Solution:**

The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)

Now slope of the line joining A and B

Equation of the line y – y_{1} = m(x – x_{1})

⇒ y – 2 = -1 (x – 0)

⇒ y – 2 = -x

⇒ x + y – 2 = 0

Point C (1, 1) will be on AB if it satisfy

1 + 1 – 2 = 0

⇒ 0 = 0

Point C lies on AB

Hence points A, C and B are collinear.

**Question 38.**

**Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:**

**(i) the co-ordinates 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 ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)

Join diagonals AC and BD which bisect each other at O.

O is mid-point of AC as well as of BD

Now co-ordinates of O will be

**Question 39.**

**In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.**

**(i) Write the co-ordinates of A.**

**(ii) Find the length of AB and AC.**

**(iii) Find the ratio in which Q divides AC.**

**(iv) Find the equation of the line AC. (2015)**

**Solution:**

In the given figure,

ABC is a triangle and BC || y-axis

AB and AC intersect the y-axis at P and Q respectively.

(i) Co-ordinates of A are (4,0).

(ii) Length of AB

**Question 40.**

**The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \(\frac { 1 }{ 2 }\). Find :**

**(i) k**

**(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)**

**Solution:**

(i) Slope of the line joining P(6, k) and Q (1 –

**Question 41.**

**A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.**

**(i) Find the coordinates of A and B.**

**(ii) Find the equation of the line through P and perpendicular to AB.**

**Solution:**

(i) Since, A lies on the x-axis,

let the coordinates of A be (x, 0).

Since B lies on the y-axis,

let the coordinates of B be (0, y).

Let m = 1 and n = 2.

Using section formula,

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E 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.