Equation of a Straight Line Class 10 RS Aggarwal Exe-14B Goyal Brothers ICSE Maths Solutions Ch-14. We Provide Step by Step Solutions / Answer of Exe-14B Questions on Conditions of Parallelism and Perpendicularity. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Equation of a Straight Line Class 10 RS Aggarwal Exe-14B Goyal Brothers ICSE Maths Solutions Ch-14
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-14 | Equation of a Straight Line |
| Writer Book | RS Aggarwal |
| Exe-14B | Conditions of Parallelism and Perpendicularity |
| Edition | 2024-2025 |
Conditions of Parallelism and Perpendicularity
Parallelism conditions: if two lines are parallel then the slope will be equal. If the slope of two parallel lines is m1 and m2, then the parallelism condition is m1 = m2.
Perpendicularity conditions: if the two lines are perpendicular then the product of their slope is -1
Exercise- 14B
( Equation of a Straight Line Class 10 RS Aggarwal Exe-14B Goyal Brothers ICSE Maths Solutions Ch-14 )
Que-1: If A(2,-3), B(-5,1), C(7,-1) and D(0,k) be four points such that AB is parallel to CD, find the value of k.
Sol: AB = m = (y2-y1)/(x2-x1)
= (1-(-3))/(-5-2) = -4/7
CD = m2 = (k-(-1))/(0-7) = (k+1)/-7
If AB || CD
Its means m2 = m1
-4/7 = (k+1)/-7
28 = 7k+7
7k = 28-7
k = 21/7 = 3.
Que-2: Show that the lines x+2y-5 = 0 and 2x+4y+9 = 0 are parallel.
Sol: Let the slope of the given lines be m1 and m2 respectively.
Then, x+2y−9 = 0
⇒ 2y = −x+9
⇒ y = (−1/2)x + (9/2)
And, 2x+4y+5 = 0
⇒ 4y = −2x−5
⇒ y = (−1/2)x − (5/4)
∴m1 = −1/2 and m2 = −1/2
Thus, m1 = m2
Hence, the given lines are parallel.
Que-3: Find the value of k for which the lines kx+2y+3 = 0 and 8x+ky-1 = 0 are parallel.
Sol: Let the slope for line kx+2y+3=0 be m₁ and for line 8x+ky-1 = 0 be m₂
Now Putting both the equations on the formula for standard line i.e. y = mx + c , we get:
⇒ kx+2y+3 = 0
⇒ y = (-kx-3)/2
⇒ y = (-kx/2) – 3/2………..(i)
thus, m₁ = -k/2
And,
⇒ 8x+ky-1 = 0
⇒ y = (-8x+1)/k
⇒ y = (-8x/k) + (1/k)
Thus, m₂ = -8k
Now,
since the lines are Parallel,
m₁ = m₂
⇒ -k/2 = -8/k
⇒ -k*k = -8*2
⇒ k = -16/-k
⇒k*k = 16
⇒k² = 16
⇒k = √16
⇒k = ±4
Que-4: If the lines 2x-by+5 = 0 and ax+3y = 2 are parallel, find the relation connecting a and b.
Sol: 2x – by + 5 = 0
by = 2x + 5
y = (2x/b) + (5/b)
Slope of this line = 2b
ax + 3y = 2
3y = −ax + 2
y = (-ax/3) + (2/3)
Slope of this line = -a/3
Since, the lines are parallel, so the slopes of the two lines are equal.
∴ 2b = -a/3
ab = −6
Que-5: Prove that the line through A(-2,6) and B(4,8) is perpendicular to the line through C(8,12) and D(4,24).
Sol: Let the slope of the line through A(-2,6) and B(4,8) be m1
m1 = (y2-y1)/(x2-x1)
m1 = (8-6)/(4+2)
m1 = 2/6 = 1/3
Let the slope of the line through C(8,12) and D(4,24) be m2
m2 = (24-12)/(4-8)
m2 = 12/-4 = -3.
Now product of slope is
m1 × m2 = (1/3) × (-3) = -1.
Thus the line are perpendicular to each other.
Que-6: If A(2,-5), B(-2,5), C(k,3) and D(1,1) be four points such that AB and CD are perpendicular to each other, find the value of k.
Sol: Given points are A(2, -5),B(-2, 5) and C(k, 3),D(1, 1)
slope = (y2-y1/x2-x1)
⇒ Slope of line AB is equal to
(5+5/-2-2) = (10/-4) = (-5/2) = -2.5
And the slope of line CD is equal to
(1-3/1-k) = (-2/1-k)
Their product must be equal to -1
the slope of line AB × Slope of line CD = -1
⇒ -2.5 x (-2/1-k) = -1
⇒ 5 = k-1
⇒ k = 6
Que-7: Prove that the lines 2x+3y+8 = 0 and 27x-18y+10 = 0 are perpendicular to each other.
Sol: In the equation of line 2x+3y+8 = 0
3y = -2x-8
y = (-2x/3) – (8/3)
m1 = (-2/3)
In the equation of line 27x-18y+10 = 0
18y = 27x+10
y = (27x/18) + (10/18)
y = (3x/2) + (5/9)
m2 = 3/2
∴ m1 × m2 = (-2/3)×(3/2)
= -1.
∴ Lines are perpendicular to each other.
Que-8: (i) If the line y = 3x+7 and 2y+px = 3 are perpendicular to each other, find the value of p.
(ii) If the straight line 3x-5y = 7 and 4x+ay+9 = 0 are perpendicular to each other, find the value of a.
Sol: (i) y = 3x + 7
Slope of this line = 3
2y + px = 3
2y = −px + 3
y = (-px/2) + (3/2)
Slope of this line = -p/2
Since, the lines are perpendicular to each other, the product of their slopes is –1.
∴ (3)(-p/2) = -1
-3p/2 = -1
p = 2/3.
(ii) 3x – 5y = 7
⇒ 5y = 3x-7
⇒ y = (3/5)x – (7/5)
⇒ Its slope = 3/5
4x + ay + 9 = 0
⇒ ay = -4x – 9
⇒ y = (-4x/a) – (9/a)
⇒ Its slope = -4/a
Since lines are perpendicular to each other,
(3/5) × (-4/a) = -1
⇒ (3/5) × (4/a) = 1
⇒ 4/a = 5/3
⇒ a = 4×(3/5) = 12/5.
Que-9: Without using Pythagoras Theorem, prove that the points A(1,3), B(3,-1) and C(-5,-5) are the vertices of a right-angled triangle.
Sol: Slope of BC
= (-5-(-1))/(-5-3) = -4/-8 = 1/2
CA = (3-(-5))/(1-(-5))
= 8/6 = 4/3
AB × BC = (1/2) × (-2)
= -1.
AB and BC is perpendicular to each other
Hence ΔABC is a right triangle.
Que-10: Without using distance formula, show that the points A(1,-2), B(3,6), C(5,10) and D(3,2) are the vertices of a parallelogram.
Sol: Slope AB = (y2-y1)/(x2-x1)
= (6-(-2))/(3-1) = 8/2 = 4
Similarly BC = (10-6)/(5-3)
= 4/2 = 2
Slope CD = (2-10)/(3-5)
= -8/-2 = 4
Slope DA = (-2-2)/(1-3)
= -4/-2 = 2.
Slope of AB and CD is equal
AB || CD and BC || DA
Hence, ABCD is a parallelogram.
Que-11: Given that A(5,4), B(-3,-2) and C(1,-8) are the vertices of a ΔABC.
Find: (i) the slope of median AD (ii) the slope of altitude BM
Sol: (i) D is the mid-point of BC
Co-ordinates = [{(1-3)/2}, {(-2-8)/2}]
= (2/2, -10/2) = (1,-5)
Now, AD = (y2-y1)/(x2-x1)
= (-5-4)/(-1-5)
= -9/-6 = 3/2.
(ii) Line AC = (-8-4)/(1-5)
= -12/-4 = 3.
BM ⊥ AC
BM = -1/slope of AC
BM = -1/3.
Que-12: (i) Find the equation of a line parallel to the line 3x+2y = 8 and passing through the point (0,1)
(ii) Find the equation of a line parallel to the line 2x+y-7 = 0 and passing through the intersection of the lines x+y-4 = 0 and 2x-y = 8.
Sol: (i) 3x + 2y = 8
2y = −3x + 8
y = (-3/2)x + (8/2)
Slope of given line = -3/2
Since the required line is parallel to given straight line.
∴ Slope of required line (m) = -3/2
Now the equation of the required line is given by:
y – y1 = m(x – x1)
⇒ y-1 = (-3/2) (x-0)
⇒ 2y – 2 = −3x
⇒ 3x + 2y = 2
(ii) From the second equation, we can rewrite it as y = 2x – 8.
Substituting this into the first equation, we have:
x – (2x – 8) – 4 = 0
x – 2x + 8 – 4 = 0
-x + 4 = 0
x = 4
Substituting the value of x back into the second equation, we have:
y = 2(4) – 8
y = 8 – 8
y = 0
Therefore, the intersection point of the lines is (4, 0).
y – y1 = m(x – x1)
Substituting the values, we have:
y – 0 = 2(x – 4)
y = 2x – 8.
Que-13: A(1,-3), B(4,2) and C(3,-2) are the vertices of a triangle.
(i) Find the co-ordinates of the centroid G of the triangle
(ii) Find the equation of the line through G and parallel to AC.
Sol: Given vertices: A( 1, 3), B(4, 2) and C(3, 2)
(i) Coordinates of the centroid G of ΔABC are given by
G = [{(-1+4+3)/3}, {(3+2-2)/3}] = (6/3, 3/3) = (2,1)
(ii) Since the line through G is parallel to AC the slope of the lines are the same
⇒ m = (y2-y1/(x2-x1) = (-2-3)/(3-(-1)) = -5/4
So, equation of the line passing through G(2, 1) and with slope -5/4 is given by
y-y1 = m(x-x1)
⇒ y-1 = (-5/4) (x-2)
⇒ 4y-4 = -5x+10
⇒ 5x+4y = 14 is the required equation.
Que-14: Find the equation of a line passing through the point P(-2,1) and parallel to the line joining the points A(4,-3) and B(-1,5).
Sol: Slope of the line = (y2-y1)/(x2-x1)
Slope of the line AB = (5+3)/(-1-4) = 8/-5
⇒ -8/5
Slope of its parallel = -8/5
The given point is p(−2, 1)
Equation of the line is y – y1 = m(x – x1)
y – 1 = (-8/5) (x+2)
5y – 5 = – 8x – 16
8x – 5y – 5 + 16 = 0
8x – 5y + 11 = 0
Que-15: Find the equation of a line passing through the origin and the perpendicular to the line y+5x = 3.
Sol: y+5x = 3
y = -5x+3
m1 = -5
m2 = 1/5 [m1×m2 = -1]
Line passing through the origin (0,0)
and perpendicular to the given line
y-0 = (1/5) (x-0)
5y = x
x-5y = 0.
Que-16: Find the equation of a line passing through the origin and parallel to the line 3x-2y+4 = 0.
Sol: Given line 3x – 2y + 4 = 0
2y = 3x + 4
⇒ y = (3/2)x + 2
⇒ Slope (m1) = 3/2 and c1 = 2
Now for the slope of parallel line (m2)
m1 = m2 = 3/2
Parallel line has the slope 32 and passes through Origin (0,0)
∴ Equation of the parallel line is(y – y1) = m (x – x1)
⇒ y – 0 = (3/2) (x – 0)
2y = 3x
3x-2y = 0
Que-17: Find the equation of the line that has x-intercept -3 and is perpendicular to the line 3x+5y = 1.
Sol: x-intercept of the line = –3
∴ Required line passes through (–3, 0) and is perpendicular to the line
3x + 5y = 1
5y = 1 – 3x
= –3x + 1
y = (-3/5)x + (1/5)
∴ Slope m = -3/5
Slope of the line perpendicular to the given line = 5/3
∴ Equation of the required line through = (–3, 0) and having slope 5/3 is given by
y-0 = (5/3)(x+3)
⇒ 3y = 5x + 15
⇒ 5x – 3y + 15 = 0
Que-18: Find the equation of the line passing through (2,4) and perpendicular to x-axis.
Sol: The line is perpendicular to x – axis and passes through (2,4)
The equation of the line perpendicular to the x – axis (y = 0) can be represented as x = c, where c is a real constant.
Now, this line passes through (2,4).
∴ c = 2
The required equation is x = 2
Que-19: Find the equation of the perpendicular dropped from the point (-1,2) onto the line joining the points (1,4) and (2,3).
Sol: Let A = (1, 4), B = (2, 3) and C = (−1, 2).
Slope of AB = (3-4)/(2-1) = -1
Slope of equation perpendicular to AB = -1/slope of AB = 1
The equation of the perpendicular drawn through C onto AB is given by:
y − y1 = m(x − x1)
y − 2 = 1(x + 1)
y − 2 = x + 1
y = x + 3.
Que-20: Find the equation of the line passing through the point of intersection of the lines 5x-8y+23 = 0 and 7x+6y-71 = 0 and perpendicular to the line 4x-2y = 3.
Sol: 7x + 6y = 71 ⇒ 28x + 24 = 284 …(1)
5x − 8y = −23 ⇒ 15x − 24y = −69 …(2)
Adding (1) and (2), we get,
43x = 215
x = 5
From (2),
8y = 5x + 23
= 25 + 23
= 48
⇒ y = 6
Thus, the required line passes through the point (5, 6).
4x − 2y = 1
2y = 4x − 1
y = 2x – (1/2)
Slope of this line = 2
Slope of the required line = -1/2
The required equation of the line is
y – y1 = m(x1,x2)
y-6 = (-1/2)(x-5)
2y − 12 = −x + 5
x + 2y = 17.
Que-21: A line through origin meets the line 2x = 3y+13 at right angle at point Q, Find the co-ordinates of Q.
Sol: Slope of parallel lines : 2x – 3y – 13 = 0 is 2/3.
Slope of perpendicular line : 2x – 3y – 13 = 0 is -3/2.
Line passes through origin means their co-ordinates = (0,0).
⇒ (y – 0) = (-3/2)(x – 0).
⇒ 2(y) = – 3x.
⇒ 2y = – 3x.
⇒ 2y + 3x = 0.
⇒ 3x + 2y = 0. – – – – – (1).
⇒ 2x – 3y = 13. – – – – – (2).
Adding equation (1) and equation (2), we get.
⇒ (3x + 2y) + (2x – 3y) = 0 + (13).
⇒ 3x + 2y + 2x – 3y = 13.
⇒ 5x – y = 13. – – – – – (3).
⇒ y = 5x – 13. – – – – – (3).
Subtracting equation (1) and equation (2), we get.
⇒ (3x + 2y) – (2x – 3y) = 0 – (13).
⇒ 3x + 2y – 2x + 3y = -13.
⇒ x + 5y = -13. – – – – – (4).
Put the value of equation (3) in equation (4), we get.
⇒ x + 5(5x – 13) = -13.
⇒ x + 25x – 65 = -13.
⇒ 26x = -13 + 65.
⇒ 26x = 52.
⇒ x = 2.
Put the value of x = 2 in equation (3), we get.
⇒ y = 5x – 13. – – – – – (3).
⇒ y = 5(2) – 13.
⇒ y = 10 – 13.
⇒ y = -3.
The coordinate of Q is (2,-3).
Que-22: Find the equation of the perpendicular from the point P(1,-2) on the line 4x-3y-5 = 0. Also, find the co-ordinates of the foot of the perpendicular.
Sol: Converting 4x – 3y – 5 = 0 in the form of y = mx + c.
⇒ 4x – 3y – 5 = 0
⇒ 3y = 4x – 5
⇒ y = (4/3)x − (5/3)
Slope of the line (m1) = 4/3.
Let the slope of the line perpendicular to 4x – 3y – 5 = 0 be m2.
Then, m1 × m2 = -1.
⇒ (4/3) × m2 = −1
⇒ m2 = −3/4.
The equation of the line having slope m2 and passing through the point (1, -2) can be given by point-slope form i.e.,
⇒ y−y1 = m(x−x1)
⇒ y−(−2) = (−3/4)(x−1)
⇒ 4(y+2) = −3(x−1)
⇒ 4y+8 = −3x+3
⇒ 3x+4y+5 = 0
For finding the coordinates of the foot of the perpendicular which is the point of
intersection of the lines
4x – 3y – 5 = 0 ….(i)
3x + 4y + 5 = 0 ….(ii)
On multiplying (i) by 4 and (ii) by 3 we get,
16x – 12y – 20 = 0 ….(iii)
9x + 12y + 15 = 0 ….(iv)
Adding (iii) and (iv) we get,
⇒ 16x – 12y – 20 + 9x + 12y + 15 = 0
⇒ 25x – 5 = 0
⇒ x = 5/25
⇒ x = 1/5.
Putting value of x in (i), we have
⇒ 4×(1/5) − 3y − 5 = 0
⇒ (4/5) − 3y − 5 = 0
⇒ 3y = (4/5)−5
⇒ 3y = (4−25)/5
⇒ 3y = −21/5
⇒ y = −7/5.
∴ Coordinates = (1/5, −7/5).
Que-23: Find the equation of the line which is perpendicular to the line (x/a)-(x/b) = 1 at the point where the given line meets y-axis.
Sol: The given line is
(x/a) – (y/b) = 1
⇒ (y/b) = (x/a) – 1
⇒ y = (b/a)x – b
Slope of this line = b/a
Slope of the required line = -1/(b/a) = -a/b
Let the required line passes through the point P(0, y).
Putting x = 0 in the equation (x/a) – (y/b) = 1 we get,
0 – (y/b) = 1
⇒ y = –b
Thus, P = (0, −b) = (x1, y1)
The equation of the required line is
y – y1 = m(x – x1)
y+b = (-a/b) (x-0)
by + b2 = −ax
ax + by + b2 = 0.
Que-24: The points B(7,3) and D(0,-4) are two opposite vertices of a rhombus ABCD. Find the equation of diagonal AC.
Sol: Rhombus ABCD with A(7, 3) and C(0, -4) as the two opposite vertices is shown in the figure below:
Slope of the line AC (m1),
= (y2−y1)/(x2−x1)
= (−4−3)/(0−7) = −7/−7 = 1.
Diagonals of rhombus bisect each other at right angles.
∴ BD is perpendicular to AC. Let slope of BD be m2.
∴ m1×m2 = −1
1 × m2 = −1
⇒ m2 = −1.
Let O be the mid-point of diagonals. It’s coordinates are given by,
= {(x1+x2)/2}, {(y1+y2)/2}
= {(7+0)/2}, {(3+(−4))/2}
= (7/2, −1/2).
Equation of BD can be given by point slope form i.e.,
⇒ y−y1 = m(x−x1)
⇒ y-(−1/2) = −1 (x−(7/2))
⇒ y+(1/2) = −x+(7/2)
⇒ (2y+1)/2 = (−2x+7)/2
⇒ 2y+1 = −2x+7
⇒ 2y+2x−6 = 0
⇒ 2(y+x−3) = 0
⇒ x+y−3 = 0.
Que-25: The points A(1,3) and C(6,8) are two opposite vertices of square ABCD. Find the equation of the diagonal BD.
Sol: the coordinates of A and C are (1,3) and (6,8)
Slope of diagonal = (y2-y1)/(x2-x1)
m1 = (8-3)/(6-1)
m1 = 5/5 = 1
Diagonal of a square is the perpendicular bisector of each other
Slope of BD = -1
O is mid-point of AC and BD
Coordinates of O are
{(1+6)/2}, {(3+8)/2}
= (7/2, 11/2)
Equation of BD will be
y-y1 = m(x-x1)
y-(11/2) = -1 (x-(7/2))
(2y-11)/2 = (-2x+7)/2
2x+2y-11-7 = 0
x+y = 9.
Que-26: A(1,4), B(3,2) and C(7,5) are the vertices of a ΔABC.
Find: (i) the co-ordinates of the centroid G of ΔABC
(ii) the equation of a line through G and parallel to AB.
Sol: (i) Co-ordinates of the centroid of triangle ABC are
= [{(1+3+7)/3}, {(4+2+5)/3}]
= [(11/3), (11/3)].
(ii) Slope of the line parallel to AB = Slope of AB = −1
Thus, the required equation of the line is
y – y1 = m(x – x1)
y-(11/3) = -1(x-(11/3))
y-(11/3) = -x+(11/3)
y+x = (11/3) + (11/3)
y+x = (11+11)/3
= 22/3
3y + 3x = 22
Que-27: A(-4,2), B(6,4) and C(2,-2) are the vertices of a ΔABC
Find: (i) the equation of median AD (ii) the equation of altitude BM (iii) the equation of right bisector of AB (iv) the co-ordinates of centroid of ΔABC
Sol: (i) D is a mid-point of BC
Coordinates of D will be
{(6+2)/2}, {(4-2)/2}
= (8/2, 2/2) = (4,1)
Equation of AB
(y-y1)/(x-x1) = (y2-y1)/(x2-x1)
(y-2)/(x+4) = (1-2)/(4+4)
(y-2)/(x+4) = -1/8
8y-16 = -x-4
x+8y-16+4
x+8y = 12.
(ii) Slope of AC
m1 = (-2-2)/(2+4) = -2/3
slope of its altitude m2 = 3/2
Equation of BM will be
(y-4) = (3/2) (x-6)
2y-8 = 3x-18
3x-2y-18+8 = 0
3x-2y-10 = 0.
(iii) Slope of AB
m1 = (2-4)/(-4-6) = 1/5
Slope of right bisector of Ab = -5
Coordinate of mid-point E
{(-4+6)/2}, {(2+4)/2}
= (1,3)
Equation will be
y-3 = -5 (x-1)
y-3 = -5x+5
5x+y-8 = 0.
(iv) Coordinate of Centroid of ΔABC will be
= {(-4+6+2)/3}, {(2+4-2)/3}
= (4/3, 4/3).
Que-28: Find the equation of the perpendicular drawn from the point P(2,3) on the line y = 3x+4. Find the co-ordinates of the foot of the perpendicular.
Sol: Given equation is y = 3x + 4 …..(i)
⇒ 3x – y + 4 = 0
Slope = 3
Equation of any line passing through the point (2, 3) is
y – 3 = m(x – 2) …..(ii)
If equation (i) is perpendicular to eq. (ii)
Then m × 3 = – 1 ……[∵m1×m2=-1]
⇒ m = -1/3
Putting the value of m in equation (ii) we get
y – 3 = (-1/3)(x-2)
⇒ 3y – 9 = – x + 2
⇒ x + 3y = 11 …..(iii)
Solving equation (i) and equation (iii) we get
3x – y = – 4
⇒ y = 3x + 4 ……(iv)
Putting the value of y in eq. (iii) we get
x + 3(3x + 4) = 11
⇒ x + 9x + 12 = 11
⇒ 10x = – 1
⇒ x = -1/10
From equation (iv) we get
y = 3(-1/10)+4
⇒ y = (-3/10)+4
⇒ y = 37/10
So the required coordinates are (-1/10, 37/10).
Que-29: A(1,2), B(2,3) and C(4,3) are the vertices of a ΔABC.
Find: (i) the equation of altitude through B
(ii) the equation of altitude through C
(iii) the co-ordinates of the orthocentre of ΔABC.
Sol: (i) A(1,2), B(2,3) and C(4,3) are the vertices of a ΔABC
Slope of AC m1 = (y2-y1)/(x2-x1)
= (3-2)/(4-1) = 1/3
Slope of BD m2 = -3
Equation of BD = y-y1 = m(x-x1)
y-3 = -3 (x-2)
y-3 = -3x+6
3x+y = 9.
(ii) Slope of AB
(3-2)/(2-1) = 1
Slope of perpendicular CE = -1
Equation of CE
y-3 = -1 (x-4)
y-3 = -x+4
x+y = 7.
(iii) After solving the equation (i) and (ii)
We shall get the coordinate of other centre H.
Subtracting (ii) from (i), we get
2x = 2
x = 1
x+y = 7
1+y = 7
y = 6
Coordinate H (1,6).
Que-30: A(1,2), B(3,-4) and C(5,-6) are the vertices of a ΔABC
Find: (i) the equation of the right bisector of BC
(ii) the equation of the right bisector of CA
(iii) the co-ordinates of the circumcentre of ΔABC.
Sol: (i) A(1,2), B(3,-4) and C(5,-6) are the vertices of a ΔABC
Slope of BC
m1 = (-6+4)/(5-3)
= -2/2 = -1
Co-ordinate of mid-point BC
{(3+5)/2}, {(-4-6)/2}
= (4,-5)
Now slope perpendicular bisector of BC m2 = 1
Equation of BC
y+5 = 1 (x-4)
y+5 = x-4
x-y = 9.
(ii) Slope AC = (-6-2)/(5-2)
= -8/4 = -2
co-ordinate of mid-point E of AC
{(1+5)/2}, {(2-6)/2}
= (3,-2)
Slope of perpendicular bisector AC = 1/2
Equation of perpendicular bisector will be
y+2 = (1/2) (x-3)
2y+4 = x-3
x-2y = 7.
(iii) After solving equation (i) and (ii)
Subtracting (i) from (ii)
y = 2
Substituting the value of y in eq (i)
x-2 = 9
x = 11.
Coordinate (11,2).
Que-31: (i) Is the line passing through the points A(-2,3) and B(4,1) perpendicular to the line 3x-y = 1?
(ii) Does the line 3x-y = 1 bisect the join of A(-2,3) and B(4,1)?
Sol: (i) Let A = (−2, 3) and B = (4, 1)
Slope of AB = m1 = (1-3)/(4+2)
= -2/6 = -1/3
Equation of line AB is
y – y1 = m1(x – x1)
y-3 = (-1/3) (x+2)
3y − 9 = −x − 2
x + 3y = 7 …(1)
Slope of the given line 3x = y + 1 is 3 = m2.
∴ m1 × m2 = −1
Yes, the line through points A and B is perpendicular to the given line.
(ii) Given line is 3x = y + 1 …(2)
Solving (1) and (2), we get,
x = 1 and y = 2
So, the two lines intersect at point P = (1, 2).
The co-ordinates of the mid-point of AB are
{(-2+4)/2, (3+1)/2} = (1,2) = P
Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.
Que-32: A line segment AB meets x-axis at A and y-axis at B. P(4,-1) divides AB in the ratio 1 : 2.
(i) Find the co-ordinates of A and B
(ii) Find the equation of a line through P and perpendicular to AB.
Sol: (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
Coordinates of P = [{(1(0)+2(x))/(1+2)}, {(1y+2(0))/(1+2)}]
⇒ (4,1) = (2×3/, y/3)
and ⇒ 2x/3 = 4 and y/3 = -1
=> x = 6 and y = -3.
So, the coordinates of A are (6,0) and that of B are (0, -3).
(ii) Slope of AB = (-3-0)/(0-6)
= -3/-6 = 1/2
=> Slope of line perpendicular to AB = m = -2
P = (4,-1)
⇒ Required equation is
y-y1 = m(x-x1)
⇒ y-(-1) = -2(x – 4)
⇒ y+1 = -2x+8
2x+y = 7.
Que-33: The vertices of a ΔABC are A(3,8), B(-1,2) and C(6,-6). Find :
(i) slope of BC (ii) Equation of a line perpendicular to BC and passing through A.
Sol: A (3, 8) , B(–1, 2) , C(6, –6)
(i) Slope of BC = (y2-y1)/(x2-x1)
= (-6-2)/(6-(-1)) = -8/7
∴ Slope of BC = – 8/7
(ii) Slope of line perpendicular to BC =-1/(-8/7) = 7/8
Required line ⇒ y – y1 = m(x – x1 )
y – 8 = (7/8) (x-3)
8y-64 = 7x-21
7x-8y+43 = 0.
Que-34: Line AB is perpendicular to CD. Co-ordinates of B, C and D are respectively (4,0), (0,-1) and (4,3). Find :
(i) stope of CD (ii) Equation of AB
Sol: (i) The coordinates of C and D are (0,-1) and (4,3) respectively.
Slope of line CD = (y2-y1)/(x2-x1)
= (3+1)/(4-0)
= 4/4 = 1.
(ii) Line AB and CD are perpendicular, so slope of line AB is
m2 = -1
Equation of line AB is
y-0 = -1 (x-4)
y = -x+4
x+y-4 = 0.
Que-35: A and B are two points on the x-axis and y-axis respectively.
(i) Write down the coordinates of A and B (ii) P is a point on AB such that AP : PB = 3 : 1. Using section formula, find the coordinates of point P. (iii) Find the equation of a line passing through P and perpendicular to AB.
Sol: (i) Coordinates of A are (4, 0)
and coordinates of B are (0, 4)
(ii) AP : PB = 3 : 1
i.e.
Coordinates of P are
[{(m1x2+m2x1)/(m1+m2)}, {(m1y2+m2y)/(1m1+m2)}]
= [{(3×0+1×4)/(3+1)}, {(3×4+1×0)/(3+1)}]
= (4/4, 12/4)
= (1, 3)
(iii) Slope of A = (y2-y1)/(x2-x1)
= (4-0)/(0-4)
= – 1
∴ Slope of the line perpendicular to AB
m = 1
Equation of line perpendicular to AB and passing through P(1, 3) is
y – y1 = m(x – x1)
⇒ y – 3 = 1(x – 1)
⇒ y – 3 = x – 1
⇒ x – y + 2 = 0
— : End of Equation of a Straight Line Class 10 RS Aggarwal Exe-14B Goyal Brothers Maths Solutions :–
Return to:- ICSE Class 10 Maths RS Aggarwal Solutions
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