# SLOPES OF PARALLEL AND PERPENDICULAR LINES

## About "Slopes of parallel and perpendicular lines"

Slopes of parallel and perpendicular lines :

If slopes of given lines are equal, then we can say that the given lines are parallel.

If two non vertical straight lines with slopes m and m , are perpendicular, then m m = –1. On the other hand, if m m = –1, then the two straight lines are perpendicular.

Note :

The straight lines x-axis and y-axis are perpendicular to each other. But, the condition m m = –1 is not true because the slope of the x-axis is zero and the slope of the y-axis is not defined.

Let us see some example problems based on the above concept.

Example 1 :

The line joining the points A (-2 , 3) and B (a , 5) is parallel to the line joining the points C (0 , 5) and D (-2 , 1). Find the value of a.

Solution :

Since the line joining the points AB and CD are parallel, the slope of those two lines will be equal.

Slope of AB :

A(-2 , 3) ==> (x, y),  B(a , 5)  ==>  (x, y)

Slope (m)  =  (y₂ - y)/(x₂ - x)

=  (5 - 3) / (a - (-2))

=  2 / (a + 2)

Slope of CD :

C(0 , 5) ==> (x, y),  D(-2 , 1) ==>  (x, y)

Slope (m)  =  (y₂ - y) / (x₂ - x)

=  (1 - 5) / (-2 - 0)

=  -4 / (-2)

=  2

Slope of AB  =  Slope of CD

2 / (a + 2)  =  2

Multiply by (a +2) on both sides

2  =  2(a + 2)

2  =  2a + 4

Subtract 4 on both sides

2 - 4  =  2a

2a  =  -2

Divide by 2 on both sides,

a  =  -1

Example 2 :

The line joining the points A(0, 5) and B (4, 2) is perpendicular to the line joining the points C (-1, -2) and D (5, b). Find the value of b.

Solution :

Since the line joining the points AB and CD are perpendicular, the product of slopes will be equal to -1.

Slope of AB :

A (0, 5) ==> (x, y),  B (4, 2) ==>  (x, y)

Slope (m)  =  (y₂ - y)/(x₂ - x)

=  (2 - 5) / (4 - 0)

=  -3 / 4

Slope of CD :

C (-1, -2) ==> (x, y),  D (5 , b) ==>  (x, y)

Slope (m)  =  (y₂ - y) / (x₂ - x)

=  (b - (-2)) / (5 - (-1))

=  (b + 2) / (5 + 1)

=  (b + 2) / 6

Slope of AB x Slope of CD  =  -1

(-3/4)  x  (b + 2)/6  =  -1

(b + 2)/8  =  1

Multiply by 8 on both sides,

b + 2  =  8

Subtract by 2 on both sides

b  =  8 - 2

b  =  6

Example 3 :

The vertices of triangle ABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

Solution :

Mid point of the side AB  =  M

mid point  =  (x₁ + x₂)/2 ,  (y₁ + y₂)/2

=  [1 + (-2)]/2 , (8 + 4)/2

=  (-1/2, 6)

Mid point of the side AC  =  N

mid point  =  (x₁ + x₂)/2 ,  (y₁ + y₂)/2

=  [1 + 8]/2 , (8 - 5)/2

=  (9/2, 3/2)

Slope of BC :

=   (y₂ - y)/(x₂ - x)

=  (-5 -4) / (8 + 2)

=  -9 / 10  ------(1)

Slope of MN :

=   (y₂ - y)/(x₂ - x)

= ((3/2) - 6) / ((9/2) + (1/2))

= (-9/2) /  (10/2)

=  -9 / 10  ------(2)

Hence, the sides MN and BC are parallel. After having gone through the stuff given above, we hope that the students would have understood "Slopes of parallel and perpendicular lines".

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