# HOW TO FIND THE EQUATION OF A LINE PARALLEL AND PERPENDICULAR

How to Find the Equation of a Line Parallel and Perpendicular ?

Here we are going to see, how to find the equation of a line parallel and perpendicular to the line joining the points.

## How to Find the Equation of a Line Parallel and Perpendicular to the Points Joining the Line - Questions

Question 1 :

Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3,-2) and R(-5, 4) .

Solution :

The required line is passing through the point P(-5, 2) and parallel to the line joining the points Q and R.

Since the line joining the points QR is parallel to the required line, the slopes will be equal.

Slope of QR  =  (y2 - y1)/(x2 - x1)

=  (4 - (-2))/(-5 - 3)

=  (4 + 2)/(-8)

=  6/(-8)

Slope of QR  =  -3/4

Slope of the required line = -3/4

Equation of the line passing through the point P.

(y - y1)  =  m(x - x1)

(y - 2)  =  (-3/4) (x - (-5))

4(y - 2)  =  -3 (x + 5)

4y - 8  =  -3x - 15

3x + 4y - 8 + 15  =  0

3x + 4y + 7  =  0

Hence the equation of the required line is 3x + 4y + 7  = 0

Question 2 :

Find the equation of a line passing through (6,–2) and perpendicular to the line joining the points (6,7) and (2,–3).

Solution :

The required line is perpendicular to the line joining the points (6, 7) and (2, -3).

Slope of the line joining the above points

m  =  (y2 - y1)/(x2 - x1)

m  =  (-3-7)/(2-6)

m  =  -10/(-4)

m  =  5/2

Slope of the required line   =  -1/(5/2)

=  -2/5

The required line is passing through the point (6, -2)

(y - y1)  =  m(x - x1)

(y - (-2))  =  (-2/5) (x - 6)

5(y + 2)  =  -2 (x - 6)

5y + 10  =  -2x + 12

2x + 5y + 10 - 12  =  0

2x + 5y - 2  =  0

After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Equation of a Line Parallel and Perpendicular".

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