FIND THE EQUATION OF A LINE PARALLEL OR PERPENDICULAR TO ANOTHER LINE

Example 1 :

Find the equation of the straight line parallel to the line

3x – y + 7 = 0

and passing through the point (1, -2).

Solution :

Write the given equation in slope-intercept form to find the slope.

3x – y + 7 = 0

-y = -3x - 7

y = 3x + 7

Comparing y = mx + b and y = 3x + 7, we get

m = 3

Slope of the given line = 3

Since the required line and given line are parallel, the slopes are equal.

Slope of the required line = 3

Equation of the required line :

y = mx + b

Substitute m = 3.

y = 3x + b ----(1)

The above line is passing through (1, -2).

-2 = 3(1) + b

-2 = 3 + b

-5 = b

Substitute b = -5 in (1).

y = 3x - 5

3x - y - 5 = 0

Example 2 :

Find the equation of the straight line perpendicular to the straight line

x – 2y + 3 = 0

and passing through the point (1, -2).

Solution :

Write the given equation in slope-intercept form to find the slope.

x – 2y + 3 = 0

-2y = -x - 3

2y = x + 3

y = (1/2)x + 3/2

Comparing y = mx + b and y = (1/2)x + 3, we get

m = 1/2

Slope of the given line = 1/2

Let k be the slope of the required line.

Since the required line and given line are perpendicular, the product of the slopes is equal to -1.

k(1/2) = -1

k = -2

Slope of the required line = -2

Equation of the required line :

y = mx + b

Substitute m = -2.

y = -2x + b ----(1)

The above line is passing through (1, -2).

-2 = 2(-1) + b

-2 = -2 + b

0 = b

Substitute b = 0 in (1).

y = -2x + 0

y = -2x

2x + y = 0

Example 3 :

Find the equation of the perpendicular bisector of the straight line segment joining the points.

(3, 4) and (-1, 2)

Solution :

Find the slope of the line joining (3, 4) and (-1, 2).

m = (y2 - y1)/(x2 - x1)

Substitute (x1, y1) = (3, 4) and (x2, y2) = (-1, 2).

m = (2 - 4)/(-1 - 3)

m = -2(-4)

m = 1/2

Let k be the slope of perpendicular bisector.

k(1/2) = -1

k = -2

Slope of the perpendicular bisector = -2.

Find the midpoint of the line joining (3, 4) and (-1, 2).

= ((x+ x2)/2, (y+ y2)/2)

= ((3 - 1))/2, (4 + 2)/2)

= (2/2, 6/2)

= (1, 3)

Perpendicular bisector is passing through (1, 3) with the slope -2.

Equation of the perpendicular bisector :

y = mx + b

Substitute m = -2.

y = -2x + b ----(1)

The above line is passing through (1, 3).

3 = -2(1) + b

3 = -2 + b

5 = b

Substitute b = 5 in (1).

y = -2x + 5

2x + y - 5 = 0

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