Question

Find an equation for the line perpendicular to the line −4x+3y=−2having the same y-intercept as −7x−7y=6

Answer 1

First we need to find slope of given line −4x+3y=−2

So convert it into slope intercept form y=mx+b

−4x+3y=−2

3y=4x−2

y=4/3x−2/3

Comparing above equation with y=mx+b gives m=4/3

Hence slope of given line is 4/3.

We need slope of perpendicular line which is always "negative reciprocal of given slope".

Negative of 4/3 is -4/3.

taking reciprocal gives -3/4

Hence slope of required perpendicular line is m=-3/4

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Now we need to find y-intercept of −7x−7y=6

y-intercept is always the y-value when x=0 so let's plug x=0 into −7x−7y=6

−7(0)−7y=6

0−7y=6

−7y=6

y=-6/7

y-intercept is also called "b" in y=mx+b

hence b=-6/7

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Now we know that slope of perpendicular line is m=-3/4 and y-intercept b=-6/7

so plug both into formula y=mx+b

We get Final answer is $y=-\frac{3}{4}x-\frac{6}{7}$