In this section, you will learn how to find the missing value if two lines are parallel or perpendicular.
Let us see how the slopes of the two lines if they are parallel or perpendicular.
If two lines are parallel, then slopes of two lines will be equal.
m1 = m2
If two lines are perpendicular, then the product of the slopes will be equal to -1.
m1 ⋅ m2 = -1
Question 1 :
Find the slope of the straight line
(i) 3x + 4y -6 = 0
Answer :
Slope (m) = - Coefficient of x/coefficient of y
= -3/4
(ii) y = 7x + 6
Answer :
y = 7 x + 6
By comparing the given equation with slope intercept form (y = mx + b), we get
Slope = 7
(iii) 4 x = 5y + 3
Answer :
4 x - 5 y - 3 = 0
Slope (m) = - coefficient of x/coefficient of y
= -4/(-5) = 4/5
Question 2 :
Show that the straight lines x + 2 y + 1 = 0 and 3x + 6y + 2 = 0 are parallel
Answer :
Since the given lines are parallel, m1 = m2
Slope of the first line x + 2 y + 1 = 0
Slope (m) = - coefficient of x/coefficient of y
m1 = -1/2 ----(1)
Slope of the second line 3x + 6y + 2 = 0
m2 = -3/6 = -1/2 ----(2)
(1) = (2)
Then the given lines are parallel.
Question 3 :
Show that the straight lines 3x – 5y + 7 = 0 and 15 x + 9y + 4 = 0 are perpendicular.
Answer :
If two lines are perpendicular then
m1 x m2 = -1
Slope of the first line 3x - 5 y + 7 = 0
Slope (m) = - coefficient of x/coefficient of y
m1 = -3/(-5)
= 3/5
Slope of the second line 15x + 9y + 4 = 0
m2 = -15/9 = -5/3
m1 ⋅ m2 = (3/5) x (-5/3) = -1
Hence the given lines are perpendicular.
Question 4 :
If the straight lines y/2 = x – p and ax + 5 = 3y are parallel, then find a.
Answer :
y/2 = x – p y = 2(x - p) y = 2x - 2p m1 = 2 -----(1) |
ax + 5 = 3y y = (a x + 5)/3 y = (a/3)x + (5/3) m2 = a/3 -----(2) |
(1) = (2)
2 = a/3
a = 6
Question 5 :
Find the value of a if the straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.
Answer :
5x – 2y – 9 = 0 m1 = -5/(-2) m1 = 5/2 -----(1) |
ay + 2x – 11 = 0 m2 = -a/2 ----(2) |
m1 ⋅ m2 = -1
(5/2) ⋅ (-a/2) = -1
-5a/4 = -1
5 a = 4
a = 4/5
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