**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.

**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 = (y_{2} - y_{1})/(x_{2} - x_{1})

= (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 - y_{1}) = m(x - x_{1})

(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 = (y_{2} - y_{1})/(x_{2} - x_{1})

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 - y_{1}) = m(x - x_{1})

(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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