EQUATION OF A LINE PARALLEL OR PERPENDICULAR TO ANOTHER LINE

Key Concept

If two lines are parallel, then the slopes are equal.

m1  =  m2

If two lines are perpendicular, then the product of the slopes is equal to -1.

m1 m2  =  -1

Example 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

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

Example 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

Example 3 :

Find the equation of the line in the xy-plane that contains the point (3, 2) and that is parallel to the line y = 4x − 1.

Solution :

Since the required line is parallel to the line y = 4x - 1,

m = 4

(y - y1)  =  m(x - x1)

(y - 2)  =  4(x - 3)

y - 2  =  4x - 12

4x - y - 12 + 2  =  0

4x - y - 10  =  0

Example 4 :

Find the equation of the line that contains the point (2, 3) and that is parallel to the line containing the points (7, 1) and (5, 6).

Solution :

Slope of the line containing the points (7, 1) and )(5, 6)

m  =  (y2 - y1)/(x2 - x1)

m  =  (6 - 1)/(5 - 7)

m  =  4/(-2)

m  =  -2

Equation of the line :

(y - y1)  =  m(x - x1)

(y - 3)  =  -2(x - 2)

y - 3  =  -2x + 4

2x + y - 3 - 4  =  0

2x + y - 7  =  0

Example 5 :

Find the equation of the line in the xy-plane that contains the point (4, 1) and that is perpendicular to the line whose equation is y = 3x + 5.

Solution :

The required line is perpendicular to the line y = 3x + 5

Slope (m)  =  3

Slope of required line  =  -1/3

Equation of the line :

(y - y1)  =  m(x - x1

Point  => (4, 1) and slope m  =  -1/3

(y - 1)  =  (-1/3)(x - 4)

3(y - 1)  =  -1(x - 4)

3y - 3  =  -x + 4

x + 3y - 3 - 4  =  0

x + 3y - 7  =  0

Example 6 :

Find a number t such that the line in the xy plane containing the points (t, 4) and (2,−1) is perpendicular to the line y = 6x − 7.

Solution :

Slope of the line containing the points (t, 4) and (2, -1)

m  =  (y2 - y1)/(x2 - x1)

m  =  (-1 - 4) / (2 - t)

m  =  -5/(2 - t)

Slope of the given line :

y = 6x − 7

m  =  6

Slope of line perpendicular to the line y = 6x - 7 is -1/6

-1/6  =  -5/(2 - t)

-1(2 - t)  =  -5(6)

-2 + t  =  -30

t  =  -30 + 2

t  =  -30

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