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)2m2, 2am) and (p2m2, 2pm)
(b) (5pq, p2q) and (5qr, qr2)
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 = (y2 – y1)/(x2 – x1)
= (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
(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
