# Frank Solutions for Chapter 13 Equation of a Straight Line Class 10 ICSE Mathematics

**Exercise 13.1**

**1. Find is the following points lie on the given line or not:**

**(i) (1, 3) on the line 2x + 3y = 11**

**(ii) (5, 3) on the line 3x – 5y + 5 = 0**

**(iii) (2, 4) on the line y = 2x – 1**

**(iv) (-1, 5) on the line 3x = 2y – 13**

**(iv) (-1, 5) on the line 3x = 2y – 13**

**Answer**

**(i)** (1, 3) on the line 2x + 3y = 11

From the question it is given that,

Point = (1, 3)

Line = 2x + 3y = 11

Now, put x = 1 and y = 3

Consider Left Hand Side (LHS) = 2x + 3y

= 2(1) + 3(3)

= 2 + 9

= 11

Right Hand Side (RHS) = 11

By comparing LHS and RHS

LHS = RHS

⇒ 11 = 11

Therefore, point lie on the given line.

**(ii)** (5, 3) on the line 3x – 5y + 5 = 0

From the question it is given that,

Point = (5, 3)

Line = 3x – 5y + 5 = 0

Now, put x = 5 and y = 3

Consider Left Hand Side (LHS) = 3x – 5y + 5

= 3(5) – 5(3) + 5

= 15 – 15 + 5

= 5

Right Hand Side (RHS) = 0

By comparing LHS and RHS

LHS ≠ RHS

⇒ 5 ≠ 0

Therefore, point does not lie on the given line.

**(iii)** (2, 4) on the line y = 2x – 1

From the question it is given that,

Point = (2, 4)

Line = y = 2x – 1

Now, put x = 2 and y = 4

Consider Left Hand Side (LHS) = 4

Right Hand Side (RHS) = 2x – 1

= 2(2) – 1

= 4 – 1

= 3

By comparing LHS and RHS

LHS ≠ RHS

4 ≠ 3

Therefore, point does not lie on the given line.

**(iv)** (-1, 5) on the line 3x = 2y – 13

From the question it is given that,

Point = (-1, 5)

Line = 3x = 2y – 15

Now, put x = -1 and y = 5

Consider Left Hand Side (LHS) = 3x

= 3(-1)

= – 3

Right Hand Side (RHS) = 2y – 13

= 2(5) – 15

= 10 – 13

= -3

By comparing LHS and RHS

LHS = RHS

⇒ – 3 = – 3

Therefore, point lie on the given line.

**(v)** (7, -2) on the line 5x + 7y = 11

From the question it is given that,

Point = (7, -2)

Line = 5x + 7y = 11

Now, put x = 7 and y = -2

Consider Left Hand Side (LHS) = 5x + 7y

= 5(7) +7(-2)

= 35 – 14

= 21

Right Hand Side (RHS) = 11

By comparing LHS and RHS

LHS ≠ RHS

⇒ 21 ≠11

Therefore, point does not lie on the given line.

**2. Find the value of m if the line 2x + 5y + 12 = 0 passes through the point (4, m)**

**Answer**

From the question it is given that,

The line 2x + 5y + 12 = 0 passes through the point (4, m)

We have to find the value of m,

So, put x = 4 and y = m

2x + 5y + 12 = 0

⇒ 2(4) + 5(m) + 12 = 0

⇒ 8 + 5m + 12 = 0

⇒ 5m + 20 = 0

⇒ 5m = – 20

⇒ m = -20/5

⇒ m = -4

Therefore, the value of m is –4.

**3. Find the value of P if the line 3y = 5x – 7 passes through the point (p, 6).**

**Answer**

From the question it is given that,

The line 3y = 5x – 7 passes through the point (p, 6)

We have to find the value of p,

So, put x = p and y = 6

3y = 5x – 7

⇒ 3(6) = 5(P) – 7

⇒ 18 = 5P – 7

⇒ 18 + 7 = 5P

⇒ 25 = 5P

⇒ P = 25/5

⇒ P = 5

Therefore, the value of P is 5.

**4. Find the value of a if the line 4x = 11 – 3y passes through the point (a, 5).**

**Answer**

From the question it is given that,

The line 4x = 11 – 3y passes through the point (a, 5)

We have to find the value of a,

So, put x = a and y = 5

4x = 11 – 3y

⇒ 4(a) = 11 – 3(5)

⇒ 4a = 11 – 15

⇒ 4a = – 4

⇒ a = -4/4

⇒ a = – 1

Therefore, the value of a is -1.

**5. The line y = 6 – 3x/2 passes through the point (r, 3). Find the value of r.**

**Answer**

From the question it is given that,

The line y = 6 – 3x/2 passes through the point (r, 3)

We have to find the value of r,

So, put x = r and y = 3

y = 6 – 3x/2

⇒ 3 = 6 – (3(r))/2

⇒ 3 = (12 – 3r)/2

⇒ 6 = 12 – 3r

⇒ 3r = 12 – 6

⇒ 3r = 6

⇒ r = 6/3

⇒ r = 2

Therefore, the value of r is 2.

**6. The line (3 + 5y)/2 = (4x – 7)/3 passes through the point (1, k). find the value of k**

**Answer**

From the question it is given that,

The line (3 + 5y)/2 = (4x – 7)/3 passes through the point (1, k)

We have to find the value of k,

So, put x = 1 and y = k

(3 + 5y)/2 = (4x – 7)/3

⇒ (3 + 5(k))/2 = (4(1) – 7)/3

⇒ 3(3 + 5k) = 2(4 – 7)

⇒ 9 + 15k = 2(- 3)

⇒ 9 + 15k = – 6

⇒ 15k = – 6 – 9

⇒ 15k = – 15

⇒ k = -15/15

⇒ k = – 1

Therefore, the value of k is – 1.

**7. The line 4x + 3y = 11 bisects the join of (6, m) and (4, 9). Find the value of m.**

**Answer**

Let us assume the point of intersection of CD and line 4x + 4y = 11 be the point Q(a, b)

From the question it is given that, line 4x + 3y = 11 bisects the line segment CD,

So, CQ: QD = 1: 1

Then, the coordinates of Q are,

Q(a, b) = Q[((6 + 4)/2), ((m – 9)/2)]

= Q[5, ((m – 9)/2)]

Since Q(a, b) lies on the line 4x + 3y = 11,

Where, x = 5, y = (m – 9)/2

4(5) + 3((m – 9)/2) = 11

⇒ 20 + (3m – 27)/2 = 11

⇒ 40 + 3m – 27 = 22

⇒ 3m + 13 = 22

⇒ 3m = 22 – 13

⇒ 3m = 9

⇒ m = 9/3

⇒ m = 3

Therefore, value of m is 3.

**8. The line 2x – 5y + 31 = 0 bisects the join of (-4, 5) and (p, 9). Find the value of p.**

**Answer**

Let us assume the point of intersection of CD and line 4x + 4y = 11 be the point Q(a, b)

From the question it is given that, line 2x – 5y + 31 = 0 bisects the line segment CD,

So, CQ: QD = 1: 1

Then, the coordinates of Q are,

Q(a, b) = Q[((-4 + P)/2), ((5 + 9)/2)]

= Q[((-4 + P)/2), 7]

Since Q(a, b) lies on the line 2x – 5y + 31 = 0,

Where, x = (-4 + P)/2, y = 7

2((-4 + P)/2) – 5(7) + 31 = 0

⇒ (-8 + 2P)/2 – 35 + 31 = 0

⇒ (-8 + 2P)/2 – 4 = 0

⇒ -8 + 2P – 8 = 0

⇒ –16 + 2P = 0

⇒ 2P = 16

⇒ P = 16/2

⇒ P = 8

Therefore, value of P is 8.

**9. The line segment formed by the points (3, 7) and (-7, Z) is bisected by the line 3x + 4y = 18. Find the value of z.**

**Answer**

Let us assume the point of intersection of CD and line 3x + 4y = 18 be the point Q(a, b)

From the question it is given that, line 3x + 4y = 18 bisects the line segment CD,

So, CQ: QD = 1: 1

Then, the coordinates of Q are,

Q(a, b) = Q[((-3 + 7)/2), ((7 + z)/2)]

= Q[-2, ((7 + z)/2)]

Since Q(a, b) lies on the line 3x + 4y = 18,

Where, x = – 2, y = (7 + z)/2

3x + 4y = 18

⇒ 3(-2) + 4((7 + z)/2) = 18

⇒ –6 + (28 + 4z)/2 = 18

⇒ –12 + 28 + 4z = 36

⇒ 16 + 4z = 36

⇒ 4z = 36 – 16

⇒ 4z = 20

⇒ z = 20/4

⇒ z = 5

Therefore, value of z is 5.

**10. The line 5x – 3y + 1 = 0 divides the join of (2, m) and (7, 9) in the ratio 2: 3. Find the value of m.**

**Answer**

Let us assume the point of intersection of CD and line 5x – 3y + 1 = 0be the point Q(a, b)

From the question it is given that, line 5x – 3y + 1 = 0divides the line segment CD are in the ratio 2: 3,

So, CQ: QD = 2: 3

So, Point C become 3(2, m) = (6, 3m)

D become 2(7, 9) = (14, 18)

Then, the coordinates of Q are,

Q(a, b) = Q[((14 + 6)/5), ((18 + 3m)/5)]

= Q[4, ((18 + 3m)/5)]

Since Q(a, b) lies on the line 5x – y + 1 = 0,

Where, x = 4, y = (18 + 3m)/5

5x – 3y + 1 = 0

⇒ 5(4) – 3((18 + 3m)/5) + 1 = 0

⇒ 20 – (54 + 9m)/5 + 1 = 0

⇒ 21 – (54 + 9m)/5 = 0

⇒ 105 – 54 – 9m = 0

⇒ 51 – 9m = 0

⇒ 9m = 51

⇒ m = 51/9

⇒ m = 17/3 **…[divide both by 3]**

Therefore, value of m is 17/3.

**11.** **The line 7x – 8y = 4 divides the join of (-8, -4) and (6, k) in the ratio 2: 5. Find the value of k.**

**Answer**

Let us assume the point of intersection of CD and line 7x – 8y = 4be the point Q(a, b)

From the question it is given that, line 7x – 8y = 4divides the line segment CD are in the ratio 2: 5,

So, CQ: QD = 2: 5

So, Point C become 5(-8, -4) = (-40, -20)

D become 2(6, k) = (12, 2k)

Then, the coordinates of Q are,

Q(a, b) = Q[((12 – 40)/7), ((2k – 20)/7)]

= Q[-4, ((2k – 20)/7)]

Since Q(a, b) lies on the line 7x – 8y = 4,

Where, x = – 4, y = (2k – 20)/7

7(-4) – 8((2k – 20)/7) = 4

⇒ – 28 – (16k – 160)/7 = 4

⇒ – 196 – 16k + 160 = 28

⇒ – 36 – 16k = 28

⇒ 16k = – 36 – 28

⇒ 16k = – 64

⇒ k = -64/16

⇒ k = -4

Therefore, value of k is – 4.

**12.** **The line 5x + 3y = 25 divides the join of (b, 4) and (5, 8) in the ratio 1: 3. Find the value of b.**

**Answer**

Let us assume the point of intersection of CD and line 5x + 3y = 25be the point Q(a, b)

From the question it is given that, line 5x + 3y = 25divides the line segment CD are in the ratio 1: 3,

So, CQ: QD = 1: 3

So, Point C become 3(b, 4) = (3b, 12)

D become 1(5, 8) = (5, 8)

Then, the coordinates of Q are,

Q(a, b) = Q[((5 + 3b)/4), ((8 + 12)/4)]

= Q[((5 + 3b)/4), 5]

Since Q(a, b) lies on the line 5x + 3y = 25,

Where, x = (5 + 3b)/4, y = 5

5((5 + 3b)/4) + 3(5) = 25

⇒ (25 + 15b)/4 + 15 = 25

⇒ 25 + 15b + 60 = 100

⇒ 15b + 85 = 100

⇒ 15b = 100 – 85

⇒ 15b = 15

⇒ b = 15/15

⇒ b = 1

Therefore, value of b is 1.

**13.** **P is a point on the line segment AB dividing it in the ratio 2: 3. If the coordinates of A and B are (-2, 3) and (8, 8), find if P lies on the line 7x – 2y = 4.**

**Answer**

From the question it is given that,

The coordinates of A and B are (-2, 3) and (8, 8)

The line segment AB dividing it in the ratio 2: 3

So, AP: PB = 2: 3

Then, A = 3(-2, 3) = (-6, 9)

B = 2(8, 8) = (16, 16)

Then, the coordinates of P are,

P(a, b) = P[((16 – 6)/5), ((16 + 9)/5)]

= P[2, 5]

Since P(a, b) lies on the line 7x – 2y = 4,

Where, x = 2, y = 5

Consider Left Hand Side (LHS) = 7x – 2y

= 7(2) – 2(5)

= 14 – 10

= 4

Right Hand Side (RHS) = 4

By comparing LHS and RHS

LHS = RHS

4 = 4

Therefore, point P(2, 5) lie on the given line 7x – 2y = 4.

**14.** **L is a point on the line segment PQ dividing it in the ratio 1: 3. If the coordinates of P and Q are (3, 7) and (11, -5) respectively, find if L lies on the line 2x + 5y = 20.**

**Answer**

From the question it is given that,

The coordinates of P and Q are (3, 7) and (11, -5) respectively

The line segment PQ dividing it in the ratio 1: 3

So, LP: LQ = 1: 3

Then, P = 3(3, 7) = (9, 21)

Q = 1(11, -5) = (11, -5)

Then, the coordinates of L are,

L(a, b) = L[((11 + 9)/4), ((- 5 + 21)/4)]

= L[5, 4]

Since L(a, b) lies on the line 2x + 5y = 20,

Where, x = 5, y = 4

Consider Left Hand Side (LHS) = 2x + 5y

= 2(5) + 5(4)

= 10 + 20

= 30

Right Hand Side (RHS) = 20

By comparing LHS and RHS

LHS ≠ RHS

30 ≠ 20

Therefore, point L(a, b) does not lie on the given line 2x + 5y = 20.

**15. ****The line segment formed by two points A(2, 3) and B(5, 6) is divided by a point in the ratio 1: 2. Find, whether the point of intersection lies on the line 3x – 4y + 5 = 0.**

**Answer**

From the question it is given that,

The coordinates of A(2, 3) and B(5, 6).

The line segment AB dividing it in the ratio 1: 2

So, AL: LB = 1: 3

Then, A = 2(2, 3) = (4, 6)

B = 1(5, 6) = (5, 6)

Then, the coordinates of L are,

L(a, b) = L[((5 + 4)/3), ((6 + 6)/3)]

= L[3, 4]

Since L(a, b) lies on the line 3x – 4y + 5 = 0,

Where, x = 3, y = 4

Consider Left Hand Side (LHS) = 3x – 4y + 5

= 3(3) – 4(4) + 5

= 9 – 16 + 5

= –2

Right Hand Side (RHS) = 0

By comparing LHS and RHS

LHS ≠ RHS

-2 ≠ 0

Therefore, point L (a, b) does not lie on the given line 3x – 4y + 5 = 0.

**Exercise 13.2 **

**1. Find the slope of a line, correct of two decimals, whose inclination is **

**(a) 60°**

**(b) 50°**

**(c) 45°**

**(d) 75°**

**(e) 30°**

**Answer**

**2. Find the inclination of a line whose gradient is **

**(a) 0.4663 **

**(b) 1.4281 **

**(c) 3.0777**

**(d) 5.6713 **

**(e) 0.5317 **

**Answer**

**(a) (2, 5) and (-1, 8) **

**(b) (3, 7) and (5, 13) **

**(c) (-5, - 1) and (-9, - 7) **

**(d) (9, - 2) and (-5, 5) **

**(e) (0, 5) and (5, 0)**

**Answer**

**4. Find the slope of a line passing through the following pairs of points: **

**(a)2m**^{2}**, 2am) and (****p ^{2}**

**m**

^{2}**, 2pm)**

**(b) (5pq, ****p ^{2}**

**q) and (5qr, qr**

^{2}**)**

**Answer**

**5. Find the slope of a line parallel to the given lines: **

**(a) 3x – 2y = 5**

**(b) x + 3y = 7 **

**(c) 5x – y = 10 **

**(d) 4x – 2y = 3 **

**(e) 5x + 2y = 11 **

**Answer**

**6. Find the value of a line perpendicular to the given lines: **

**(a) 2x – 3y = 4 **

**(b) 5x + 2y – 9 = 0 **

**(c) 3x + 4y = 13 **

**(d) x – 4y = 8 **

**(e) 9x – 3y = 5 **

**Answer**

**7. Find the slope of a line parallel to AB, if the coordinates of A and B are (3, - 1) and (-7, 5) respectively. **

**Answer**

**8. Find the slope of a line parallel to MN, if the coordinates of M and N are (4, 9) and (-2, 3) respectively. **

**Answer**

**9. Find the slope of a line parallel to PQ, if the coordinates of P and Q are (-11, 3) and (7, 13) respectively. **

**Answer**

Slope of line PQ = (y_{2} – y_{1})/(x_{2} – x_{1})

= (13 + 3)/(7 – 11)

= 16/-4

= -4

Slope of line parallel to PQ = Slope of PQ

= -4

**10. Find x if the slope of the line passing through the points (x, 9) and (12, 6) is -1/3. **

**Answer**

Slope of line AB

⇒ -1/3 = (6 – 9)/(12 – x)

⇒ x – 12 = - 9

⇒ x = 3

**11. Find m if the slope of the line passing through the point (-7, 5) and (2, m) is 1/3. **

**Answer**

**12. Find p if the slope of the line passing through the point (-2, 5) and (p, 2p + 1) is 1. **

**Answer**

**Answer**

**14. Find the slope and inclination of the line passing through (- 5, 7) and (7, - 5). **

**Answer**

**15. Find the value of a line parallel to the following lines: **

**(a) x = y/2 – 5 **

**(b) x = 3y/2 + 2 **

**(c) 3x/4 + 5y/2 = 7 **

**(d) x/4 + y/3 = 1 **

**(e) 2x/5 + y/3 = 2 **

**Answer**

**16. Find the slope of a line perpendicular to the following lines:**

**(a) x/2 + y/3 = 4/3 **

**(b) x – 3y/2 + 1 = 0 **

**(c) 3x/4 – y = 5 **

**(d) 3x – 5y = 9 **

**(e) 4x + y = 7 **

**Answer**

**17. Without using distance formula, show that the points A(12, 8), B (- 2, 6) and C (6, 0) from a right-angled triangle. **

**Answer**

**18. Without distance formula, show that the points P (2, 1), Q (- 5, - 1), R (1, 5) and S (- 2, - 1) from a parallelogram. **

**Answer**

**19. Without distance formula, show that the points A (5, 8), B (4, 4), C(0, 0) and D(1, 9) form a rhombus. **

**Answer**

**20. Without distance formula, show that the points M (1, 7), N (4, 8), O (5, 5) and P (2, 4) form a square. **

**Answer**

**Exercise 13.3 **

**1. Find the slope and the y-intercept of the follow lines: **

**(a) 5x – 2y = 6 **

**(b) 3x + y = 7 **

**(c) 4y = 5x – 8 **

**(d) 2x + 3y = 12 **

**(e) x – 2 = (5 – 3y)/2 **

**Answer**

**2. Find the equation of a line whose slope and y-intercept are **

**(a) m = -6/5, c = 3 **

**(b) m = 2/3, c = 2 **

**(c) m = -1/2, c = 5 **

**(d) m = - 3, c = - 1**

**(e) m = 2, c = - 5 **

**Answer**

Equation of line with slope and y-intercept

**3. Find the equation of a line passing through (2, 5) and making an angle of 30°with the positive direction of the x-axis. **

**Answer**

**4. Find the equation of a line passing through (3, 7) and making an angle of 60° with the negative direction of the x-axis. **

**Answer**

**Answer**

**6. Find the equation of a line passing through (2, 9) and parallel to the line 3x + 4y = 11. **

**Answer**

**7. Find the equation of a line passing through (-5, - 1) and perpendicular to the line 3x + y = 9. **

**Answer**

**8. Find the equation of the perpendicular bisector of AB if the coordinates of A and B are (-2, 6) and (4, - 6). **

**Answer**

**9. Find the equation of a line perpendicular to the join of A (3, 5) and B(-1, 7) if it passes through the midpoint of AB. **

**Answer**

**10. Find the equation of a line passing through the intersection of x + 3y = 6 and 2x – 3y = 12 and parallel to the line 5x + 2y = 10 **

**Answer**

**11. Find the equation of a line passing through the intersection of x + 2y + 1 = 0 and 2x – 3y = 12 and perpendicular to the line 2x + 3y = 9. **

**Answer**

**12. Find the equation of a line passing through the intersection of x/10 + y/5 = 14 and x/8 + y/6 = 15 and perpendicular to the line x – 2y = 5. **

**Answer**

**13. The lines px + 5y + 7 = 0 and 2y = 5x – 6 are perpendicular to each other. Find p. **

**Answer**

**14. The lines 3x – 2y + 4 = 0 and 3x + my + 6 = - are parallel to each other. Find m. **

**Answer**

**15. Find the relation connecting p and q, if the lines py = 2x + 5 and qx + 3y = 2 are parallel to each other. **

**Answer**

**16. Find the relation connecting a and b, if the lines ay = 2x + 4 and 4y + bx = 2 are perpendicular to each other. **

**Answer**

**17. P (5, 3), Q (-4, 7) and R (8, 3) are the vertices of a triangle. Find the equation of the median of the triangle from p. **

**Answer**

**18. A (8, 5), B (-2, 1) and C (5, 4) are the vertices of a triangle. Find the equation of the median of the triangle through C. **

**Answer**

**19. ABCD is a rhombus. The coordinates of A and C (3, 7) and (9, 15). Find the equation of BD. **

**Answer**

**20. ABCD is a square. The coordinates of B and D are (-3, 7) and (5, - 1) respectively. Find the equation of AC. **

**Answer**

**21. The coordinates of two points P and Q are (0, 4) and (3, 7) respectively. Find: **

**(i) the gradient of PQ. **

**(ii) the equation of PQ. **

**(iii) the coordinates of the point where the line AB intersects the x-axis. **

**Answer**

**22. X(4, 9), Y(-5, 4) and Z(7, - 4) are the vertices of a triangle. Find the equation of the altitude of the triangle through X. **

**Answer**