Write the equation of a line perpendicular to 6x 5y 3 tha

Write the equation of a line perpendicular to 6x - 5y = - 3 that passes through the pant (- 6, 3). The equation of the line is y = (Simplify your answer Type your answer in slope-intercept form Use integers or fractions for any numbers in the expression.)

Solution

Given that

6x - 5y = -3   

First we have to convert this line in slope intercept form y = mx + b for finding slope

- 5y = -3 - 6x

5y = 3 + 6x

5y = 6x + 3

y = (6x + 3)/5

y = (6/5)x + (3/5)

Hence ,

Slope m = 6/5

i.e Slope of first line m₁ = 6/5

Perpendicular lines have negative reciprocal slopes.

m₁.m₂ = -1

(6/5) .m₂ = -1

m₂ = -1 .(5/6)

m₂ = -5/6

Slope of second line m₂ = -5/6

y = mx + b

y = (-5/6)x + b

Substitute (-6,3 ) for finding b

3 = (-5/6).(-6) + b

3 = 5 + b

3 - 5 = b

b = -2

Hence ,

The equation of line is ,

y = mx + b

y = (-5/6)x + (-2)

y = (-5/6)x - 2

Therefore ,

The equation of line is ,   y = (-5/6)x - 2