SLOPES OF PARALLEL AND PERPENDICULAR LINES

Slopes of Parallel Lines

If two non vertical straight lines are parallel, then slopes of the lines will be equal. 

If two non vertical straight lines with slopes m1 and m2, are parallel, then

m1  =  m2

On the other hand, if the slopes of two straight lines are equal, then the lines will be parallel. 

Slopes of Perpendicular Lines

If two non vertical straight lines are perpendicular, then the product of slopes of the lines will be equal to '-1'. 

If two non vertical straight lines with slopes m1 and m2, are perpendicular, then

m1m2  =  –1

On the other hand, if the product of slopes of two straight lines is equal to '-1', then the lines will be perpendicular. 

Note :

The straight lines x-axis and y-axis are perpendicular to each other. But, the condition m1m2 = –1 is not true. Because the slope of the x-axis is zero and the slope of the y-axis is not defined. 

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) ==> (x1, y1),  B(a, 5)  ==>  (x2, y2)

Slope (m1)  =  (y2 - y1)/(x2 - x1)

  =  (5 - 3)/(a - (-2))

  =  2/(a + 2)

Slope of CD :

C(0 , 5) ==> (x1, y1),  D(-2, 1) ==> (x2, y2)

Slope (m2)  =  (y2 - y1)/(x2 - x1)

  =  (1 - 5)/(-2 - 0)

  =  -4/(-2)

  =  2

Slope of AB  =  Slope of CD

2/(a + 2)  =  2

Multiply each side by (a + 2). 

2  =  2(a + 2)

2  =  2a + 4

Subtract 4 from each side. 

2 - 4  =  2a 

2a  =  -2

Divide each side by 2. 

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) ==> (x1, y1),  B (4, 2) ==>  (x2, y2)

Slope (m1)  =  (y2 - y1)/(x2 - x1)

  =  (2 - 5)/(4 - 0)

  =  -3 / 4

Slope of CD :

C (-1, -2) ==> (x1, y1),  D (5 , b) ==>  (x2, y2)

Slope (m2)  =  (y2 - y1)/(x2 - x1)

  =  (b - (-2)) / (5 - (-1))

  =  (b + 2) / (5 + 1)

  =  (b + 2) / 6

(Slope of AB)  (Slope of CD)  =  -1

(-3/4) ⋅ (b + 2)/6  =  -1

(b + 2)/8  =  1

Multiply each side by 8.

b + 2  =  8

Subtract 2 from each side. 

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 - y1)/(x2 - x1)

  =  (-5 - 4)/(8 + 2)

  =  -9/10  ------(1)

Slope of MN :

=   (y2 - y1)/(x2 - x1)

  = ((3/2) - 6)/((9/2) + (1/2)) 

  = (-9/2)/(10/2)

  =  -9/10  ------(2)

From (1) and (2),

slope of BC  =  slope of MN

So, the sides MN and BC are parallel.

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