HOW TO FIND THE SLOPE FROM AN EQUATION

Find the slope of the following straight lines :

Example 1 :

5y - 3 = 0

Solution :

5y - 3 = 0 

5y = 3

y = 3/5

This is the equation of an horizontal line passing through the value 3/5 on the y-axis.

We know that the slope of an horizontal line is always zero.

Hence the slope of the given line is 0.

Example 2 :

7x - 3/17 = 0

Solution :

7x - 3/17 = 0

7x = 3/17

x = 3/119

This is the equation of a vertical line passing through the value 3/119 on the x-axis.

We know that the slope of a vertical line is always undefined.

Hence the slope of the given line is undefined.

Example 3 :

Find the slope of the line which is

(i) parallel to y = 0.7x - 11

(ii) perpendicular to the line x = -11

Solution :

Part (i) :

y = 0.7x  - 11

The given equation of the line is in slope-intercept form.

Comparing y = mx + b and y = 0.7x - 11, we get

m = 0.7

Slope of the given line = 0.7

If the lines are parallel, then the slopes must be equal.

Then, the slope of the line parallel to y = 0.7x - 11 is also 0.7.

Part (ii) :

x = -11

This is the equation of a vertical line passing through the value -11 on the x-axis.

We know that the slope of a vertical line is always undefined.

So, the slope of the line x + 11 = 0 is undefined.

Let k be the slope of the line perpendicular to x + 11 = 0.

If the lines are perpendicular, then the product of the slopes must be equal to -1.

k x undefined = -1

= -1/undefined

= 0

Then, the slope of the line perpendicular to x = -11 is zero.

How to Check If the Given Lines are Parallel or Perpendicular

Example 4 :

Check whether the following two lines are parallel.

x/3 + y/4 + 1/7 = 0

2x/3 + y/2 + 1/10 = 0

Solution :

x/3 + y/4 + 1/7 = 0 ----(1)

2x/3 + y/2 + 1/10 = 0 ----(2)

Let m₁ and m₂ be the slopes of the lines (1) and (2) respectively.

Find the slope of line (1) :

x/3 + y/4 + 1/7 = 0

y/4 = -x/3 - 1/7

y = (-4/3)x - 4/7

Comparing y = mx + b and y = (-4/3)x - 4/7, we get

m = -4/3

Slope of line (1) = -4/3

m₁ = -4/3

Find the slope of line (2) :

2x/3 + y/2 + 1/10 = 0

y/2 = -2x/3 - 1/10

y = (-4/3)x - 2/10

y = (-4/3)x - 1/5

Comparing y = mx + b and y = (-4/3)x - 1/5, we get

m = -4/3

Slope of line (2) = -4/3

m₂ = -4/3

m₁ = m₂

Since the slopes are equal, the lines are parallel.

Example 5 :

Check whether the following two lines are perpendicular.

5x + 23y + 14 = 0

23x − 5y + 9 = 0

Solution :

Let m₁ and m₂ be the slopes of the lines (1) and (2) respectively.

Find the slope of line (1) :

5x + 23y + 14 = 0

23y = -5x - 14

y = (-5/23)x - 14/23

Comparing y = mx + b and y = (-5/23)x - 14/23, we get

m = -5/23

Slope of line (1) = -5/23

m₁ = -5/23

Find the slope of line (2) :

23x - 5y + 9 = 0

-5y = -23x - 9

5y = 23x + 9

y = (23/5) + 9/5

Comparing y = mx + b and y = (23/5)x + 9/5, we get

m = 23/5

Slope of line (2) = 23/5

m₂ = 23/5

m₁ ⋅ m₂ = (-5/23)(23/5)

m₁ ⋅ m₂ = -1

Since the product of the slopes is equal to -1, the lines are perpendicular.

Example 6 :

If the following two lines are perpendicular, then find the value of p.

12y = -(p + 3)x + 12

12x - 7y = 16

Solution :

12y = -(p + 3)x + 12 ----(1)

12x - 7y = 16 ----(2)

Find the slope of line (1) :

12y = -(p + 3)x + 12

y = [-(p + 3)/12]x + 12/12

y = [-(p + 3)/12]x + 1

Comparing y = mx + b and y = [-(p + 3)/12]x + 1, we get

m = -(p + 3)/12

Slope of line (1) = -(p + 3)/23

Find the slope of line (2) :

12x - 7y = 16

-7y = -12x + 16

7y = 12x - 16

y = (12/7)x - 16/7

Comparing y = mx + b and y = (12/7)x - 16/7, we get

m = 12/7

Slope of line (2) = 12/7

Since the lines are perpendicular, product of the slopes is equal to -1.

[-(p + 3)/12][12/7] = -1

-(p + 3)/7 =  -1

p + 3 = 7

p = 4

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