ML Aggarwal ICSE Solutions Class 10 Math 12th Chapter Equation of Straight Line
ML Aggarwal ICSE Solutions Class 10 Chapter 12 Equation of Straight LineExercise 12.1
(1) Find the slope of the line whose inclination is
(1) 45°
Solution:
Given that, θ = 45°
Then, Slope = tan 45° = 1
(ii) 30°
Solution:
Given that, θ = 30°
Then, Slope = tan 30° = 1/ √3.
(2) Find the inclination of a line whose gradient is
(i) 1
Solution:
Given that,
tan θ = 1
= θ = tan (1)
θ = 45°
(ii) √3
Solution:
Given that,
tan θ = √3
θ = tan-1 (√3)
θ = 60°
(iii) 1/ √3
Solution:
Given that,
tan θ = 1/ √3
θ = tan-1 (1/√3)
θ = 30°
(3) Find the equation of a straight line parallel to X – axis which is at a distance
(i) 2 units above it
Solution:
We have, Equation of a line parallel to
X – axis y = a
= y = 2
Y – 2 = 0 is the required equation of a straight line.
(ii) 3 units below it
Solution:
We have,
Equation of a line parallel to X – axis is y = a
= y = -3
Y + 3 = 0 is the required equation of a straight line.
(4) Find the equation of a straight line parallel to Y – axis which is a distance of
(i) 3 units to the right
Solution:
We have,
The equation of a line parallel to Y – axis is x = a
= x = 3
= x – 3 = 0 is the required of a straight line.
(ii) 2 units to the left
Solution:
We have,
The equation of a line parallel to Y – axis is x = a
= x = -2
= x + 2 = 0 is the required equation of a straight line.
(5) Find the equation of a straight line parallel to Y – axis and passing through the point (-3,5).
Solution:
We have,
The equation of a straight line parallel to Y – axis is x = a.
But, here the line is passing through the point (-3, 5).
Hence, the required equation of a straight line parallel to Y – axis and passing through the point (-3, 5) is found to be
X = -3 = x + 3 = 0
(6) Find the equation of a line whose (i) Slope = 3, (ii) Y – intercept = -5
Solution:
Given that, Slope (m) = 3 and Y – intercept (c) = -5
We have, equation of a line in Slope – intercept form is
Y = mx + C
m – Slope
Y – intercept = c
Y = m x + c
Y = 3 x – 5
3x – y – 5 = 0 is the required equation of a line.
(ii) Slope = -2/7, Y – intercept = 3
Solution:
Given that,
Slope (m) = -2/7 and
Y – intercept (c) = 3
We have, equation of a straight line in slope – intercept form is Y = m x + c
Y = -2/7 x + 3
7y = -2x + 21
2x + 7y – 21 = 0 is the required equation of a line.
iii.)
(7) Find the slope & Y – intercept of the following lines:
(i) x – 2y – 1 = 0
Solution:
Given equation of line is
X – 2y – 1 = 0
X = 2y + 1
2y = x – 1
Y = 1/2 x – 1/2 —- (1)
On complaining equation (1) with y = m x + c
= m = 1/2 and C = -1/2
Thus, here {Slope (m) = 1/2 & Y – intercept (c) = – 1/2}
(ii) 4x – 5y – 9 = 0
Solution:
Given equation of a line is
4x – 5y – 9 = 0
4x = 5y + 9
5y = 4x – 9
Y = 4/5 x – 9/5 —- (b)
On complaining equation (1) with Y = m x + c
= m = 4/5 and c = -9/5
{Thus, here Slope (m) = 4/5 & y intercept (c) = -9/5}
(iii) 3x + 5y + 7 = 0
Solution:
Given equation of line is 3x + 5y + 7 = 0
3x = – 5y – 7
5y = -3x -7
Y = -3/5 x – 7/5 —- (1)
On complaining equation (1) with
Y = m x + c
= m = -3/5 & c = -7/5
Thus, here Slope (m) = -3/5
Y – intercept (c) = -7/5
(iv) x/3 + 7/4 = 1
Solution:
Given equation of a line is x/3 + 7/4 = 1
4x + 3y/12 = 1
4x + 3y = 12
3y = 12 – 4x
Y = -4/3 × +4 —- (1)
On complaining equation (1) with y = m x + c
= m = -4/3 & c = 4
Thus, here Slope (m) = -4/3 & Y – intercept (c) = 4
(8) The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ enlaces with the positive direction of X – axis.
Solution:
Given that, the equation of the line PQ is 3y – 3x + 7 = 0
i.e. 3y = 3x – 7
y = x – 7/3 —- (1)
On complaining equation (1) with y = m x + c
(i) We get Slope of the line PQ (m) = 1
(ii) And tan θ = 1
θ = tan-1 (1)
θ = 45°
(9) The given fig. represents the line Y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the X – axis. Hence, determine θ.
Solution:
Given equations of lines are y = x + 1 — (1) and y = √3x – 1 —– (2)
On complaining equation (1) & (2) with y = m x + c
We get,
For (1): Slope (m) = 1, y – intercept (c) = 1
For (2): Slope (m) = √3, y – intercept (c) = – 1
Here, tan θ = √3
θ = tan-1 (√3) = 60°
θ = 60°
In fig, the triangle is formed by given two lines & X – axis.
Thus, exterior angle = Sum of interior opposite angles
60° = θ + 45°
θ = 60° – 45°
θ = 15° is the required angle.
(10) Find the value of P, given that line y/2 = x – p passes through the point (-4, 4).
Solution:
Given that, the line y/2 = x – p passes through the point (-4, 4). i.e. y/2 = x –p —- (1)
Put x = – 4 and y = 4 in equation (1)
4/2 = – 4 – P
2 = – 4 – P
P = – 4 – 2
P = – 6 is the required value.
(11) Given that (a, 2a) lies on the line y/2 = 3 x – 6 find a.
Solution:
Given that, The point (a, 2a) lies on the line y/2 = 3x – 6 —(1)
Then, Put x = a and y = 2a in equation (1)
2a/2 = 3a – 6
a = 3a – 6
6 = 2a
a = 6/2 = 6/2
= a = 3 is the required value of a.
(12) The graph of the equation y = m x + c passes through the points (1, 4) and (-2, -5). Determine the values of m and C.
Solution:
Given equation of line y = m x + c passes through the points (1, 4) and (-2, -5).
Put x = 1 & y = 4 in y = m x + c
4 = m + c —- (1)
Put x = – 2 and y = – 5 in y = m x + c
= – 5 = – 2m + c —– (2)
(1) × 2 = 8 = 2m + 2c
(2) = -5 = -2m + 2c
3 = 3c
C = 1 Put in (1) = 4 = m + c
4 = m + 1
m = 3
Thus, the values of m and c are found to be m = 3 & c = 1.
(13) Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y – axis.
Solution:
Given that, A line passes through the point (2, – 5) and makes an intercept of -3 on the y – axis.
We have, the equation of line in slope – intercept form is given by, y = m x + c = y = m x -3 — (1)
and put x = 2 & y = – 5 in y = m x + c ∵ c = -3
-5 = 2m + c
2m = -5 – c = – 5 + 3
2m = – 2
m = -1
Thus, the required equation of line is found to be
Y = – x – 3 i.e. x + y + 3 = 0
(14) Find the equation of a straight line passing through (-1, 2) and whose slope is 2/5.
Solution:
Given that,
A line passes through the point (-1, 2) & having slope 2/5.
= x1 = – 1, y = 2 and m = 2/5
We have, the equation of a line m slope point form as
Y – y1 = m (x – x1)
Y – 2 = 2/5 (x + 1)
5y – 10 = 2x + 2
2x – 5y + 12 = 0 is the required equation of line.
(15) Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).
Solution:
Given that, a line is having inclination of 60° & passes through the point (0, – 3).
Through the point (0, – 3).
= x1 = 0, y1 = – 3 and θ = 60°
Slope (m) = tan θ = tan 60° = √3
We have, equation of a line in slope – point form is found to be
Y – Y1 = m (x – x1)
Y + 3 = √3 (x – 0)
√3x – y – 3 = 0 is the required equation of line.
(16) Find the gradient of a line passing through the following pairs of points.
(i) (0, – 2), (3, 4)
Solution:
Here (x, y1) ≡ (0, – 2)
And (x2, y2) ≡ (3, 4)
Then Slope/ gradient of a line is given by
m = y2 – y1/ x2 – x1 = 4 + 2/ 3 – 0 = 6/3
m = 2
(ii) (3, – 7), (- 1, 8)
Solution:
Here, (x1, y1) ≡ (3, – 7)
and (x2, y2) ≡ (- 1, 8)
Then slope/ gradient of a line is given by
m = y2 – y1/ x2 – x1 = 8 + 7/ – 1 – 3 = 15/ – 4
m = -15/4
(17) The co – ordinates of two points E and F are (0, 4) and (3, 7) respectively. Find (i) the gradient of EF
(ii) the equation of EF
(iii) the Co – ordinates of the point where the line EF meter sects the X – axis.
Solution:
Given that, the co – ordinates of two points E & F are (0, 4) & (3, 7) respectively.
(i) Gradient of EF is given by m = y2 – y1/ x2 – x1 = 7 – 4/ 3 – 0 = 3/3 = 1
m = 1
(ii) We have, equation of a line in Slope – point form is given by
Y – y1 = m (x – x1)
= y – 7 = 1 (x – 3)
Y – 7 = x – 3
X – y + 4 = 0 is the required equation of a line.
(iii) If the line x – y + 4 = 0 intersects X – axis then y = 0
Put y = 0 in x – y + 4 = 0
X + 4 = 0
X = – 4
Thus, the co – ordinates of the point where the line EF intersects X – axis is found to be x = – 4 & y = 0
(18) Find the intercept made by the line 2x – 3y + 12 = 0 on the co – ordinates axis.
Solution:
To get intercept made on X – axis put y = 0 in 2x – 3y + 12 = 0
= 2x + 12 = 0
2x = – 12
X = – 6
To get intercept made on y – axis put x = 0 in 2x – 3y + 12 = 0
= 0 – 3y + 12 = 0
3y = 12
y = 4
(19) Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence show that the points P, Q, & R (11, 4) are collinear.
Solution:
Given points are P (5, 1), Q (1, – 1), R (11, 4)
To find equation of a line passing through points P & Q:
We have, Slope (m) = y2 – y1/ x2 – x1 = -1 – 1/ 1 – 5 = -2/ – 4 = 1/2
m = 1/2
Now, equation of a line passing through points P & Q and having slope 1/2 is given by,
Y – y1 = m (x – x1)
Y – 1 = 1/2 (x – 5)
2y – 2 = x – 5
X – 2y – 3 = 0 is the required equation.
To show points P, Q, & R are collinear:
Put x = 11 & y = 4 in x – 2y – 3 = 0
11 – 8 – 3 = 0
11 – 11 = 0
0 = 0
Thus, the points P (5, 1), Q (1, – 1) & R (11, 4) are collinear.
Hence Proved.
(22) 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) the equation of the diagonal BD.
Solution:
Given that,
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) & D (2, – 4).
(i) Let us consider ‘o’ is the point of intersection of diagonals of the parallelogram ABCD.
= (x, y) = (5 + 2/ 2, 8 – 4/ 2) = (3.5, 2)
For line AC:
3.5 = (x + 4)/2 & 2 = (y + 7)/2
7 = x + 4 & 4 = y + 7
X = 3 & y = – 3
Thus, the Co – ordinates of point A are found to be A (3, – 3).
(ii) The Slope of a line BD is given by
Slope (m) = – 4 – 8/ 2 – 5 = – 12/ – 3 = 4
m = 4
We have, equation of a line BD in Slope – point form is
Y – y1 = m (x – x1)
Y – 8 = 4 (x – 5)
4x – y – 20 + 8 = 0
4x – y – 12 = 0 is the required equation of the diagonal BD.
(23) In △ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.
Solution:
Given that,
In △ABC, A (3, 5), B (7, 8) and C (1, – 10).
In fig AD is the median.
= Point D is the mid – point of side BC.
Then, By mid – Point formula,
D (x, y) ≡ (7 + 1/2, 8 – 10/2) ≡ (4, – 1)
Now, Slope of the median AD is given by
m = y2 – y1/ x2 – x1 = 5 + 1/ 3 – 4 = 6/ – 1 = -6
m = – 6
Now, equation of a line AD in Slope – point form is given by,
Y – y1 = m (x – x1)
Y + 1 = -6 (x – 4)
Y + 1 = -6x + 24
6x + y – 23 = 0 is the required equation of median through Point A.
(24) Find the equation of a line passing through the point (-2, 3) and having X – intercept 4 units.
Solution:
Given that,
A line passes through the point (-2, 3) & having X – intercept 4 units.
= The Co – ordinates of the point be (4, 0).
Now, Slope of the line passing through two points (-2, 3) and (4, 0) is given by
m = y2 – y1/ x2 – x1 = 0 – 3/ 4 + 2 = – 3/6 = -1/2
= m = -1/2
Now, equation of a line in slope – point form is given by
Y – y1 = m (x – x1)
Y – 0 = -1/2 (x – 4)
2y = – x + 4
X + 2y = 4 is the required equation of a line.
(25) Find the equation of the line whose X – intercept y – intercept is -4.
Solution:
Given that,
The X – intercept of a line = 6 and
The y – intercept of a line = – 4 = C = – 4
Thus, the required line passes through the points (6,0) & (0, – 4).
Slope (m) = y2 – y1/ x2 – x1 = – 4 – 0/ 0 – 6 = – 4/ – 6 = 2/3
= m = 2/3
Now, equation of a line in slope – intercept form is given by
Y = m x + c
Y = 2/3 x – 4
3y = 2x – 12
2x – 3y – 12 = 0 is the required equation of a line.
(27) Find the equation of the line passing through point (1, 4) & intersecting the line x – 2y – 11 = 0 on the Y – axis.
Solution:
Given that,
A line passes through the point (1, 4) & inter sects the line
X – 2y – 11 = 0 on the Y – axis.
= x = 0 put in x – 2y – 11 = 0
= y = 11/2
Now, slope of the line passing through points (1, 4) & (0, – 1/2) is m = y2 – y1/ x2 – x1 = -11/2 – 4/ 0 – 1 = 19/2
Thus, the equation of a line in slope – point form is given by,
Y – y1 = m (x – x1)
Y + 11/2 = 19/2 (x – 0)
2y + 11 = 19 x
19 x – 2y – 11 = 0 is the required equation of a line.
(28) Find the equation of the straight line containing the point (3, 2) & making positive equal intercepts on axes.
Solution:
- Let us consider the required line passes through the point P (3, 2).
- Also, it passes through X – axis at point A (x, 0) and Y – axis at point B (0, y).
From given condition & fig, OA = OB
= x = y
The slope of the line is given by,
m = y2 – y1/x2 – x1 = 0 – y/ x – 0 = – x/ – x = -1
Thus, equation of a line slope – point form is given by,
Y – y1 = m (x – x1)
Y – 2 = – 1 (x – 3)
Y – 2 = – x + 3
X + y – 5 =0 is the required equation of line.
(30) A and B are two points on the X – axis and Y – axis respectively. P (2, – 3) is the mid – point of AB. Find
(i) the co – ordinates of A and B
(ii) the Slope of line AB
(iii) the equation of line AB
Solution:
Given that, A and B are two points on the X – axis & Y – axis respectively & P (2, – 3) is the mid – point of AB.
(i) Let us consider the co – ordinates be A (x, 0) & B (0, y).
= 2 = (x + 0)/2 & – 3 = (0 + y)/2
X = 4 & y = – 6
(ii) The slope of line AB is given by
m = y2 – y1/ x2 – x1 = – 6 – 0/ 0 – 4 = – 6/ – 4 = 3/2
= m = 3/2
(iii) Now, equation of line AB in Slope – point form is given by,
Y – y1 = m (x – x1)
Y – (-3) = 3/2 (x – 2)
Y + 3 = 3/2 (x – 2)
2y + 6 = 3 x – 6
3 x – 2y – 12 = 0 is the required equation of line.
(32) The line through point P (5, 3) intersects y – axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co – ordinates of Q
Solution:
Given that,
The line passes through point P (5, 3) & intersects y – axis at point Q as shown.
(i) Slope of line (m) = tan 45° = 1
m = 1
(ii) Equation of line PQ in Slope – point form is given by
Y – y1 = m (x – x1)
Y – 3 = 1 (x – 5)
Y = x – 2 or x – y – 2 = 0 is the required equation of line.
(iii) Put x = 0 in equation x – y – 2 = 0
= y = – 2
Thus, Co – ordinates of point Q are found to be Q (0, – 2).
(35) Find the equation of a straight line passing through the origin & through the point of intersection of the lines 5x + 7y = 3 and 2x – 3y = 7.
Solution:
Given that,
A line passes through the origin & through the point of intersection of the lines 5x + 7y = 3 & 2x – 3y = 7.
5x + 7y = 3 —- (1)
2x – 3y = 7 — (2)
(1) × 3 = 15 x + 21 y = 9
(2) × 7 = 14 x – 21 y = 49
29 x = 58
x = 2 put in (1)
5 (2) + 7y = 3
10 + 7y = 3
7y = – 7
Y = – 1
Thus, the required line passes through the points (0, 0) & (2, – 1).
Then, Slope of line (m) = y2 – y1/ x2 – x1 = (0 + 1)/ (0 – 2) = -1/2
= m = -1/2
Thus, equation of line is given by,
Y – y1 = m (x – x1)
Y- 0 = -1/2 (x – 0)
2y = – x
= x + 2y = 0 is the required equation of line.
Exercise 12.2
(1) State which one of the following is true:
The Straight lines y = 3x – 5 and 2y = 4x + 7 are
(i) Parallel
(ii) Perpendicular
(iii) neither parallel nor Perpendicular
Solution:
Given equation of lines are y = 3x – 5 — (1)
2y = 4x + 7 —- (2)
Slope of line (1) = m1 =3
Slope of line (2) = m2 = 2
Here, m1 ≠ m2 and m1 m2 ≠ – 1
Hence, the required lines are neither Parallel nor perpendicular.
(2) If 6x + 5y – 7 = 0 and 2 p x + 5y + 1 = 0 are parallel lines, find the value of P.
Solution:
Given lines are 6x + 5y – 7 = 0 —- (1)
i.e. 5y = – 6x + 7
y = – 6/5 x + 7/5
i.e. y = m x + c
Here, the slope of line (1) is found to be m1 = – 6/5
Now, Given equation of line are 2px + 5y + 1 = 0 — (2)
i.e. 5y = -2px -1
y = -2/5px x – 1/5
i.e. y = m x + c
Here, the slope of line (2) is found to be m2 = -2/5 P
But, given lines (1) & (2) are parallel.
= m1 = m2
-6/5 = -2/5 P
= 6/2 = P
P = 3 is the required value of P.
(3) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.
Solution:
Given equation of lines are 2x – by + 5 = 0 — (1)
and ax + 3y – 2 = 0 —— (2)
(1) = – by = – 2x – 5
Y = 2/b x + 5/b
i.e. y = m x + c
Here, Slope of line (1) is found to be m1 = 2/b
(2) = 3y = – ax + 2
Y = – a/3 x + 2/3
i.e. y = m x + c
Here, slope of line (2) is found to be m2 = – a/3
But, given that, the lines (1) & (2) are Parallel.
= m1 = m2
2/b = – a/3
= 6 = – a b
ab = – 6 is the required relation between a and b.
(4) If the straight lines 3x – 5y = 7 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a.
Solution:
Given equations of line are
3x – 5y = 7 —– (1)
and 4x + ay + 9 = 0 —– (2)
(1) = 5y = 3x – 7
Y = 3x/5 – 7/5
i.e. y = m x + c
Here, the slope of line (1) is found to be m1 = 3/5
(2) = 4x + ay + 9 = 0
ay = – 4x – 9
y = – 4/ a x – 9/ a
i.e. Y = m x + c
Here, the slope of line (2) is found to be m2 = – 4/a
Given that, lines (1) & (2) are perpendicular.
m1 m2 = – 1
(3/5) (- 4/a) = – 1
12/5a = 1
5a = 12
a = 12/5 is the required value of a.
(5) If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation betn a and b.
Solution:
Given equation of lines are 3x + by + 5 = 0 —– (1)
and ax – 5y + = 0 ——- (2)
(1) = by = – 3x – 5
Y = -3/b x – 5/b
i.e. y = m x + c
Here, slope of line (1) is found to be m1 = 3/b
(2) = 5y = ax + 7
Y = a/5 x + 7/5
i.e. y = m x + c
Here, slope of line (2) is found to be m2 = a/5
Given that, the lines (1) & (2) are perpendicular to each other.
= m1 m2 = – 1
(-3/b) (a/5) = – 1
3a/5b = 1
= 3a = 5b is the required relation between a and b.
(6) Is the line through (- 2, 3) and (4, 1) perpendicular to the line 3x = y + 1? Does the line 3x = y + 1 bisect the join of (-2, 3) & (4, 1)
Solution:
Given line is passing through the points (-2, 3) & (4, 1).
Then slope of a line (m1) = y2 – y1/ x2 – x1 = 1 – 3/ 4 + 2 = – 2/6 = – 1/3
m1 = -1/3
And slope of the line 3x = y + 1
i.e. y = 3x – 1
Slope (m2) = 3
m2 = 3
Here, m1 m2 = – 1
Thus, the given lines are perpendicular to each other.
Now, Co – ordinates of the mid – point of the line joining the points (-2, 3) and (4, 1) is found to be
(-2 + 4/2, 3 + 1/2) ≡ (1, 2)
Put x = 1 & y = 2 in equation 3x = y + 1
= 3 (1) = 2 + 1
3 = 3
Thus, the line 3x = y + 1 bisects the line which joins the points (-2, 3) & (4, 1).
(8) The line through A (-2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
Given line is passing through two points A (-2, 3) & B (4, b).
Then slope of a line (m1) = y2 – y1/ x2 – x1 = b – 3/ 4 + 2 = b -3/6
And the line with equation 2x – 4y = 5 is perpendicular to above line through points A and B.
Here, 4y = 2 x – 5
Y = 1/2 x – 5/4
i.e. y = m x + c
= m2 = 1/2
Since, given two lines are perpendicular to each other.
= m1 m2 = – 1
(b – 3/6) (1/2) = – 1
b – 3 = – 12
b = – 12 + 3
b = – 9 is the required value of b.
(9) Find the equation of a line, which has the Y – intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the co – ordinates of the point where it cuts the X – axis.
Solution:
Given that, The Y – intercept of required line is C = 4
Which is parallel to the line 2x – 3y – 7 = 0 —– (1)
3y = 2x – 7
Y = 2/3 x – 7/3
i.e. y = m x + c
m = 2/3 is the slope of line — (1)
If two lines are parallel then their slopes are also equal.
Thus, the required line is having slope 2/3 and Y – intercept as 4 whose equation is given by
Y = m x + c
Y = 2/3 x + 4
= 3y = 2x + 12
i.e. 2x – 3y + 12 = 0 is the required equation of line. —- (2)
When the line (2) cuts the X – axis then y = 0
Put y = 0 in (2)
2x + 12 = 0
2x = – 12
x = – 6
Thus, the co – ordinates of the required point are (- 6, 0).
(10) Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with Y – intercept – 3.
Solution:
Given equation of a line is 2x + 5y + 7 = 0 —– (1)
5y = – 2x – 7
Y = -2/5 x – 7/5
i.e. y = m x + c
Here, slope of line (1) is found to be m1 = – 2/5
And the line with y – intercept – 3 is perpendicular to (1)
= m1 m2 = – 1
m2 = – 1/m1 = – 1/ (- 2/ 5) = 5/2
m2 = 5/2
Now, equation of a line with slope 5/2 and y – intercept – 3 is given by
Y = m x + c
Y = 5/2 x – 3
2y = 5x – 6
5x – 2y – 6 = 0 is the required equation of line.
(11) Find the equation of a straight line perpendicular to the line 3x – 4y + 12 = 0 and having same y – intercept as 2x – y + 5 = 0.
Solution:
Given equation of a line is 3x – 4y + 12 = 0 —- (1)
Slope of line (1) = 4y = 3x +12
Y = 3/4 x + 3
i.e. y = m x + c
Slope of line (1) = m1 = 3/4
Line (1) is perpendicular to the line having Y – intercept as 2x – y + 5 = 0.
Here, y = 2x + 5
i.e. y = m x + c
i.e. Y – intercept is found to be c = 5
But, given lines are perpendicular to each other.
m1 m2 = – 1
3/4 (m2) = – 1
= m2 = – 4/3
Thus, required equation of line with slope – 4/3 & y – intercept 5 is given by,
Y = m x + c
Y = – 4/3 x + 5
3y = – 4x + 15
i.e. 4x + 3y – 15 = 0 is the required equation of line.
(13) Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (-1, -2).
Solution:
Given equation of a line is 3x + 8y = 12 —- (1)
(1) = 8y = – 3x + 12
Y = – 3/8 x + 6/4 = – 3/8 x + 3/2
i.e. y = m x + c
Then, slope of line (1) is found to be m1 = – 3/8
The required line is perpendicular to line (1).
= m1 m2 = – 1
(- 3/8) m2 = – 1
m2 = 8/3
Now, equation of a line with slope 8/3 & passing through point (-1, -2) is given by
(15) Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid – point of the line segment joining the points (2, 7) and (-4, 1).
Solution: Given equation of the line is 2x + 5y – 7 = 0 —(1)
(1) = 5y = – 2x + 7
Y = – 2/5 x + 7/5 i.e. y = m x + c
Here, Slope of line (1) is found to be m1 = – 2/5
The required line is parallel to line (1) = m2 = – 2/5
Also, the required line passes through the mid – point of the line segment joining the points (2, 7) & (-4, 1).
Then, (x, y) ≡ (2 – 4/2, 7 + 1/2) ≡ (-2/2, 8/2) ≡ (-1, 4) is the mid – Point
Now, equation of the line passing through (-1, 4) & with slope – 2/5 is given by,
Y – y1 = m (x – x1)
Y – 4 = (- 2/5) (x + 1)
5y – 20 = – 2x – 2
2x + 5y – 18 = 0 is the required equation.
(17) Find the equation of a Straight line passing through the intersection of 2x + 5y – 4 = 0 with X – axis and parallel to the line 3x – 7y + 8 = 0
Solution: Given equation of line is 3x – 7y + 8 = 0 — (1)
7y = 3x + 8
Y = 3/7 x + 8/7 i.e. y = m x + c
Then, slope of a line (1) is found to be m1 = 3/7
The required line is parallel to line (1) = m2 = 3/7
Also, the required line passes through the intersection of line 2x + 5y – 4 = 0 & X – axis.
Put y = 0 in 2x + 5y – 4 = 0
2x – 4 = 0
2x = 4
X = 2
Thus, required line is passing through the point (2, 0 & having slope 3/7.
Thus, equation of a line in slope – point form is given by,
Y – y1 = m (x – x1)
Y – 0 = 3/7 (x – 2)
7y = 3x – 6
3x – 7y – 6 = 0 is the required equation of line.
(19) Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also, find the co – ordinates of the foot of perpendicular.
Solution: Given equation of line is 4x – 3y – 5 = 0 —- (1)
= 3y = 4x – 5
Y = 4/3 x – 5/3 i.e. y = m x + c
Hence, slope of line (1) is found to be m1 = 4/3
Then, the slope of line perpendicular to (1) is m2 = – 3/4
Now, equation of a line passing through point (1,- 2) and having slope – 3/4 is given by,
Y – y1 = m (x – x1)
Y + 2 = – 3/4 (x – 1)
4y + 8 = – 3x + 3
3x + 4y + 5 =0 — (2)
(1) × 4 = 16x – 12y = 20
(2) x 3 = 9x + 12y = – 15
25 x = 5
X = 1/5 put in (1)
4 (1/5) – 3y = 5
4/5 – 3y = 5
4 – 15y = 25
4 – 25 = 15y
Y = – 21/15
Thus, the co – ordinates of foot of perpendicular is found to be (1/5, – 21/15).
(20) Prove that the line through (0,0) and (2,3) is parallel to the line through (2, – 2) and (6,4).
Solution: Given line is passing through points (0,0) & (2,3).
Then slope of line = m1 = y2 – y1/x2 – x1 = 3 – 0/2 – 0 = 2/3 m1 = 2/3
Again line which is passing through points (2, – 2) & (6, 4).
Then slope of line = m2 = y2 – y1/x2 – x1 = 4 + 2/ 6 – 2 = 6/4 = 3/2
Here, m1 = m2
Thus, given two lines are parallel to each other.
Hence Proved.
(21) Prove that the line through (- 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).
Solution: Given line is passing through Points (- 2, 6) and (4, 8).
Then slope of line is found to be m1 = y2 – y1/ x2 – x1 = 8 – 6/ 4 + 2 = 2/6 = 1/3 m1 = 1/3
And, the line is passing through points (8, 12) & (4, 24)
Then slope of line is found to be
m2 = y2 – y1/ x2 – x1 = 24 – 12/ 4 – 8 = 12/- 4 = – 3
= m2 = 3
Here, m1 m2 = – 1
Thus, given two lines are perpendicular to each other.
Hence Proved.
(22) Show that the triangle formed by the points A (1, 3), B (3, – 1) and c (-5, -5) is a right angled triangle by using slopes.
Solution:
Given that,
The three vertices of a triangle are A(1, 3), B (3, -1) and c(-5, -5)
Then, slope of line AB = m1 = y2 – y1/x2 – x1 = -1 -3/3 – 1 = -4/2 = 2
m1 = -2
and slope of line BC = m2 = y2 – y1/x2 – x1 = -5 + 1/-5 – 3 = -4/-8 = 1/2
m2 = 1/2
Thus, m1 m2 = -1
Hence, required triangle is a right angled triangle.
Since line AB is perpendicular to line BC.
Hence Proved.
(23) Find the equation of line through the point (-1, 3) and parallel to the line joining the points (0, 2) and (4, 5).
Solution: Given line is passing through the points (0, 2) and (4, 5).
Then slope of line (m1) = y2 – y1/x2 – x1 = 5 + 2/4 – 0 = 7/4
But the required line is parallel to above line.
Thus, slope of required line is found to be
m2 = 7/14
Then equation of line passing through point (-1, 3) and having slope 7/4 is given by,
Y – y1 = m (x – x1)
Y – 3 = 7/4 (x + 1)
4y – 12 = 7x + 7
7x – 4y + 19 = 0 is the required equation of line.
(25) Find the equation of the line through (0, 3) and perpendicular to the line joining the points (-3, 2) and (3, 1).
Solution: Given line is passing through the points (-3, 2) & (9, 1).
Then, slope of line = m1 = y2 – y1/x2 – x1 = 1 – 2/ 9 + 3 = -1/12
m1 = -1/12
But, required line is perpendicular to above line.
= m2 = -1/m1 = 12 = m2 = 12
Now, the required line is passing through point (0, -3) & having slope 12 whose equation is given by
Y – y1 = m (x – x1)
Y + 3 = 12 (x – 0)
Y + 3 = 12 x
12 x – y – 3 = 0 is the required equation of line.
(27) The vertices of a triangle are A (10, 4), B (4, -9) and c (-2, -1). Find the equation of the attitude through A. (The perpendicular drawn from a vertex of a triangle to the opposite side is called attitude).
Solution:
Given that,
The vertices of a triangle are A (10, 4), B (4, -9) and c (-2, -1).
Then slope of line BC = m1 = y2 – y1/x2 – x1 = -1 +9/-2 – 4 = 8/-6 = -4/3
Let us consider, the slope of the attitude from the point A (10, 4) to BC be m2.
Then, m1 x m2 = -1
-4/3 (m2) = -1
m2 = 3/4
Thus, equation of line passing through point (10, 4) & having slope 3/4 is given by
Y – y1 = m (x – x1)
Y – 4 = 3/4 (x – 10)
4y – 16 = 3x – 30
3x – 4y – 14 = 0 is the required equation of line.
(29) Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, -6).
Solution:
Given line is passing through the points (1, 2) & (5, -6).
Slope of the line = m1 = y2 – y1/x2 – x1 = -6 – 2/5 – 1 = -8/4 = -2
Let us consider m2 be the slope of right bisector of the line segment.
= m1 m2 = -1
m2 = -1/m1 = -1/-2 = 1/2
m2 = 1/2
Then, the mid – point of the line joining points (1, 2) and (5, -6) is found to be
(1 + 5/2, 2 -6/2) ≡ (6/2, -4/2) ≡ (3, -2)
Thus, equation of line passing through point (3, -2) and with slope 1/2.
Then, y – y1 = m (x – x1)
Y + 2 = 1/2 (x – 3)
2y + 4 = x -3
X – 2y – 7 = 0 is the required equation of line.
(31) The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.
Solution:
Given that,
The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD.
Then slope of line BD = m1 = y2 – y1/x2 – x1 = 8 – 3/6 – 1 = 5/5 = 1
m1 = 1
We have, the diagonal AC is a perpendicular bisector of diagonal BD.
= m1 × m2 = -1
m2 = -1/m1 = -1/1 = -1
= m2 = -1
Then, equation of line passing through point (7/2, 11/2) and having slope -1 is given by
Y – y1 = m (x – x1)
Y – 11/2 = -1 (x – 7/2)
2y – 11 = – 2x + 7
2x + 2y – 18 = 0 i.e. x + y – 9 = 0 is the required equation.
(32) ABCD is a rhombus. The co – ordinates of A and C are (3, 6) and (- 1, 2) respectively. Write down the equation of BD.
Solution:
Given that, ABCD is a rhombus with co – ordinates of vertices A (3, 6) & C (-1, 2) respectively.
Then, slope of AC = m1 = y2 – y1/x2 – x1 = 2 -6/-1 -3 = -4/-4 = 1
m1 = 1
Already we know that,
The diagonals of a rhombs bisect each other at right angle.
= diagonals BD and AC are perpendicular to each other.
If m2 is the slope of BD = m2 = -1
Now, co – ordinates of mid – point of AC is found to be
(3 – 1/2, 6 + 2/2) ≡ (2/2, 8/2) ≡ (1, 4)
Thus, equation of line passing through point (1, 4) and having slope – 1 is given by
Y – y1 = m (x – x1)
Y – 4 = -1 (x – 1)
= y – 4 = -x +1
X + y – 5 = 0 is the required equation of line.