# Find an equation for the line perpendicular to the line 5x-20y=3 having the same y-intercept as -7x-4y=8

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### 2 Answers

The first equation 5x-20y=3 solved for y=mx+b is `y=1/4x-3/20` Since the equation has a slope (m) of 1/4 the perpendicular slope would be -1/m or -4. So my answer equation has to be y=-4x+b.

The second equation -7x-4y=8 tells me the y-intercept solved for y=mx+b it's `y=-7/4 x-2` b is the y-intercept and it's -2.

So I need a slope (m) of -4, and a y-intercept (b) of -2. That gives me an equation `y=-4x-2`

let us write given equation -7x-4y=8 in intercept form as

`x/(8/-7)+y/(8/-4)=1`

`x/(-8/7)+y/(-2)=1` (i)

y intercept of equation (i) is -2

Rewrite equation 5x-20y=3 ,in slope intercept form

-20y=-5x+3

`y= (-5x)/(-20)+3/(-20)`

`y=(1/4)x-3/20` (ii)

slope of line (ii) is (1/4)

required line to be perpendicular to (ii) ,let slope of required line be m . So

m.(1/4)=-1

m=-4

So slope of required line is -4 and intercept is -2

So equation of line will be

y=-4x-2

(y=mx+c)