POINT SLOPE FORM EQUATION OF A LINE

We will find the equation of a straight line passing through a given point A(x1, y1) and having the slope m.

Let P(x, y) be any point other than A on the given line.

Slope of the line joining A(x1, y1) and P(x, y) is given by

m = (y - y1)/(x - x1)

m(x - x1) = (y - y1)

Hence, equation of line in point-slope form : 

y - y1 = m(x - x1)

Example 1 :

Find the point-slope form equation of a line passing through the point (1, 2) with slope -4.

Solution :

Given : Point = (1, 2) and slope m = -4.

Equation of line in point-slope form :

y - y1 = m(x - x1)

Substitute (x1 , y1) = (1 , 2) and m = -4.

y - 2 = -4(x - 1)

Example 2 :

Find the point-slope form equation of a line passing through the point (-2, 3) with slope 1/3.

Solution :

Given : Point = (-2, 3) and slope m = 1/3.

Equation of line in point-slope form :

y - y1 = m(x - x1)

Substitute (x1 , y1) = (-2 , 3) and m = 1/3.

y - 3 = (1/3)[x - (-2)]

Example 3 :

Find the point-slope form equation of the straight line passing through the point (-5, -4) with slope 2/3.

Solution :

Given : Point = (-5, -4) and slope m = 2/3.

Equation of line in point-slope form :

y - y1 = m(x - x1)

Substitute (x1 , y1) = (-5 , -4) and m = 2/3.

y - (-4) = (2/3)[x - (-5)]

Example 4 :

Find the point-slope form equation of the straight line passing through the point (1, 2) and parallel to the line whose equation is x + 2y + 3  =  0.

Solution :

Write the equation of the line 'x + 2y + 3 = 0' in slope intercept form.

x + 2y + 3 = 0

2y = -x - 3

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

So, slope of the given line is -1/2.

Because the required line is parallel to the given line, the slopes are equal.

Then, slope of the required line is -1/2.

The required line is passing through (1, 2) with slope -1/2.

Equation of line in point-slope form :

y - y1 = m(x - x1)

Substitute (x1 , y1) = (1, 2) and m = -1/2.

y - 2 = (-1/2)(x - 1)

Example 5 :

Find the point-slope form equation of the straight line passing through the point (-2, 3) and perpendicular to the line whose equation is x - 2y - 6  =  0.

Solution :

Write the equation of the line 'x - 2y - 6 = 0' in slope intercept form.

x - 2y - 6 = 0

-2y = -x + 6

2y = x - 6

y = (1/2)x - 3

So, slope of the given line is 1/2.

Because the required line is perpendicular to the given line, product of the slopes is equal to -1.

Let 'm' be the slope of the required line.

m x (1/2) = -1

m/2 = -1

m = -2

So, the required line is passing through (-2, 3) with slope -2.

Equation of line in point-slope form :

y - y1 = m(x - x1)

Substitute (x1 , y1) = (-2, 3) and m = -2.

y - 3 = -2[x - (-2)]

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