Science > Mathematics > Coordinate Geometry > Straight Lines > Condition for Parallelism and Perpendicularity
In this article, we shall study the Condition for parallelism and the condition of perpendicularity of lines.
Condition for Parallelism:
If two lines are parallel, then their slopes are equal. Let slopes of the two parallel lines be m1 and m2, then by the condition of parallelism,
m1 = m2.
Condition for Perpendicularity:
If two lines are perpendicular, then the product of their slopes is – 1. Let slopes of the two perpendicular lines be m1 and m2, then by the condition of parallelism,
m1 x m2. = – 1
Example – 01:
Find the value of k so that the line joining (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6).
Solution:
Let A(3, k), B(2, 7), and C(-1, 4), D(0, 6) be the given points
Given line AB ∥ line CD
∴ Slope of line AB = Slope of line CD
∴ (7 – k)/(2 – 3) = (6 – 4)/(0 + 1)
∴ (7 – k)/(-1) = 2/1
∴ (7 – k) = 2
∴ k = 7 + 2
∴ k = 9
Example – 02:
Find the value of k so that the line joining (2, k) and (-3, 5) is parallel to the line through (-1, 5) and (0, 7).
Solution:
Let A(2, k), B(-3, 5), and C(-1, 5), D(0, 7) be the given points
Given line AB ∥ line CD
∴ Slope of line AB = Slope of line CD
∴ (5 – k)/(-3 – 2) = (7 – 5)/(0 + 1)
∴ (5 – k)/(-5) = 2/1
∴ (5 – k) = 2(-5)
∴ 5 – k = – 10
∴ k = 5 + 10
∴ k = 15
Example – 03:
Find the value of k so that seg AB is parallel to Seg CD where A(-3, 1), B(4, 3), C(k, 3), and D(2, 5).
Solution:
Given points are A(-3, 1), B(4, 3), C(k, 3), and D(2, 5)
Given seg AB ∥ seg CD
∴ Slope of seg AB = Slope of seg CD
∴ (3 – 1)/(4 + 3)) = (5 – 3)/(2 – k)
∴ 2/7 = 2/(2 – k)
∴ 1/7 = 1/(2 – k)
∴ 2 – k = 7
∴ k = 2 – 7
∴ k = – 5
Example – 04:
A(0, 9), B(1, 11), C(3, 13) and D(7, k) are the four points if AB ⊥ CD then find k
Solution:
Given points are A(0, 9), B(1, 11), and C(3, 13), D(7, k)
Slope of AB = (11 – 9)/(1 – 0) = 2/1 = 2
Slope of CD = (k – 13)/(7 – 3) = (k – 13)/4
Given line AB ⊥ line CD
∴ Slope of line AB x Slope of line CD = -1
∴ 2 x ((k – 13)/4) = – 1
∴ (k – 13)/2 = – 1
∴ k – 13 = – 2
∴ k = – 2 + 13
∴ k = 11
Example – 05:
A(0, 9), B(1, 5), C(-2, 7) and D(4, k) are the four points if AB ⊥ CD then find k.
Solution:
Given points are A(0, 9), B(1, 5), C(-2, 7) and D(4, k)
Slope of AB = (5 – 9)/(1 – 0) = – 4/1 = – 4
Slope of CD = (k – 7)/(4 + 2) = (k – 7)/6
Given line AB ⊥ line CD
∴ Slope of line AB x Slope of line CD = -1
∴ – 4 x ((k – 7)/6) = – 1
∴ – 4(k – 7) = – 6
∴ – 4k + 28 = – 6
∴ – 4k = – 34
∴ k = 34/4
∴ k = 17/2
Example – 06:
A(3, 2), B(-2, 7), C(-5, -4) and D(-2, k) are the four points if AB ⊥ CD then find k
Solution:
Given points are A(3, 2), B(-2, 7), C(-5, -4) and D(-2, k)
Slope of AB = (7 – 2)/(-2 – 3) = 5/(-5) = – 1
Slope of CD = (k + 4)/(-2 + 5) = (k + 4)/3
Given line AB ⊥ line CD
∴ Slope of line AB x Slope of line CD = -1
∴ – 1 x ((k + 4)/3) = – 1
∴ k + 4 = 3
∴ k = – 1
Example – 07:
The line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Solution:
Let A(-2, 6), B(4, 8), and C(8, 12), D(x , 24) be the given points
Slope of AB = (8 – 6)/(4 + 2) = 2/6 = 1/3
Slope of CD = (24 – 12)/(x – 8) = 12/(x – 8)
Given line AB ⊥ line CD
∴ Slope of line AB x Slope of line CD = -1
∴ (1/3 x (12/(x – 8)) = – 1
∴ 12/(x – 8)= – 3
∴ 12 = – 3x + 24∴ 3x = 12
∴ x = 4
Example – 08:
If A(6, 4) and B(2, 12) are two given points then find the slope of the line (i) parallel to AB and (ii) perpendicular to AB
Solution:
Given A(6, 4) and B(2, 12)
Slope of AB = (12 – 4)/(2 – 6) = (8)/(- 4) = – 2
Let l be the line parallel to line AB
As lines are parallel their slopes are equal
∴ Slope of line l = slope of line AB = – 2
Let m be the line perpendicular to line AB
As lines are perpendicular to each other, product of their slopes is -1
∴ Slope of line m = – 1/slope of line AB = – (1/- 2) = 1/2
The slope of the line parallel to AB is – 2 and that of the line perpendicular to AB is 1/2
Example – 09:
2x + 3y + 7 = 0 and 4x + 6y + 2 = 0are two straight lines. Are they parallel to each other?
Solution:
Equation of first line is 2x + 3y + 7 = 0
Slope of first line = – (coefficient of x / Coefficient of y)
Slope of first line = – (2/3) = -2/3 ………. (1)
Equation of second line is 4x + 6y + 2 = 0
Slope of second line = – (coefficient of x / Coefficient of y)
Slope of second line = – (4/6) = -2/3 ………. (2)
From equation (1) and (2) we have
Slope of first line = Slope of second line
Hence the two lines are parallel to each other.
Example – 10:
Show that the lines 3x – 2y + 6 = 0 and 2x + 3y – 1 = 0are perpendicular to each other?
Solution:
Equation of first line is 3x – 2y + 6 = 0
Slope of first line = – (coefficient of x / Coefficient of y)
Slope of first line = – (3/-2) = 3/2 ………. (1)
Equation of second line is 2x + 3y – 1 = 0
Slope of second line = – (coefficient of x / Coefficient of y)
Slope of second line = – (2/3) = -2/3 ………. (2)
Slope of first line x Slope of second line = (3/2) x (-2/3) = -1
Hence the two lines are perpendicular to each other.
Example – 11:
Show that the lines 5x + 6y -1 = 0 and 6x – 5y + 3 = 0are perpendicular to each other?
Solution:
Equation of first line is 5x + 6y -1 = 0
Slope of first line = – (coefficient of x / Coefficient of y)
Slope of first line = – (5/6) = – 5/6 ………. (1)
Equation of second line is 6x – 5y + 3 = 0
Slope of second line = – (coefficient of x / Coefficient of y)
Slope of second line = – (6/-5) = 6/5 ………. (2)
Slope of first line x Slope of second line = (- 5/6) x (6/5) = -1
Hence the two lines are perpendicular to each other.
Example – 12:
Find value of ‘k’, if the lines kx – 6y = 9 and 6x + 5y = 13 are perpendicular to each other.
Solution:
Equation of first line is kx – 6y – 9 = 0
Slope of first line = – (coefficient of x / Coefficient of y)
Slope of first line = – (k/-6) = k/6 ………. (1)
Equation of second line is 6x + 5y – 13 = 0
Slope of second line = – (coefficient of x / Coefficient of y)
Slope of second line = – (6/5) = – 6/5 ………. (2)
Given the two lines are perpendicular to each other
Slope of first line x Slope of second line = -1
(k/6) x (-6/5) = -1
k/5 = 1
k = 5
Example – 13:
Find value of ‘k’, if the lines 3x + 4py + 8 = 0 and 3py – 9x + 10 = 0 are perpendicular to each other.
Solution:
Equation of first line is 3x + 4py + 8 = 0
Slope of first line = – (coefficient of x / Coefficient of y)
Slope of first line = – (3/4p) ………. (1)
Equation of second line is – 9x + 3py + 10 = 0
Slope of second line = – (coefficient of x / Coefficient of y)
Slope of second line = – (-9/3p) = 9/3p ………. (2)
Given the two lines are perpendicular to each other
Slope of first line x Slope of second line = -1
(- 3/4p) x (9/3p) = -1
9/4p2 = 1
p2 = 9/4
p = ± 3/2