Find the equation of the line perpendicular to the line 3x 5

Find the equation of the line perpendicular to the line 3x +5y = -2 and passing through the point (2, -6). Write your answer in slope-intercept form y = mx + b.

Solution

First we have to determine the slope of the line given

3x + 5y = -2 [ convert the line to y = mx + b]

5y = -2 - 3x

y = -2/5 -3x/5

thus b = -2/5 and m = -3/5

Two lines are perpendicular if their slopes m1 . m2 = -1

Assume slope of the given line is m1 and

slope of new line passing through point (2, -6) is m2

thus m1 . m2 = -1

m2 = -1/m1 [insert the slope obtained above -3/5]

m2 = -1/(-3/5)

m2 = 5/3

now we have determined the slope of the new line and the intercept is same as given line

m2 = 5/3 and b = -2/5

now substitute the values in the slope-intercept form y = mx + b

y = (5/3) .x + (-2/5)

y = 5x/3 - 2/5 [multiply the left term with 5 and and right term with 3 both numerator and denominator to have same denominators]

y = 25x/15 - 6/15

y = (25x - 6)/15

15y = 25x - 6

0 = 25x - 6 - 15y

25x -15y -6 = 0

thus the equation of the line perpendicular to the line 3x + 5y = -2 and passsing through the point (2,-6) is

25x - 15y = 6