How do you write an equation of a line passing through (2, -8), perpendicular to 4x+5y=7?

1 Answer
Mar 14, 2017

y=5/4x-21/2

Explanation:

First, you would want to get only y on one side of the given equation (for converting the equation to point slope form), first by subtracting 4x from both sides:

5y=7-4x or 5y=-4x+7

Then divide by 5:

y=7/5-4/5x or y=-4/5x+7/5

Product of slopes of two perpendicular lines is -1, so you know that the slope of the line perpendicular to given line would be

(-1)/(-4/5)=1xx5/4=5/4

Now, make your new equation in point-slope format:

y=mx+b

y=5/4x+b

Now, to figure out the y-intercept (b), you can substitute 2 and -8 into x and y, respectively i.e.

(-8)=5/4xx2+b

Multiply 5/4 and 2:

(-8)=10/4+b or (-8)=5/2+b

Adding -5/2 to both sides:

(-8)-5/2=b

Make -8 and 5/2 have a common denominator:

-16/2-5/2=b

-21/2=b

Then you can put that into the point-slope formula and you're done!

y=5/4x-21/2
graph{(y-5/4x+21/2)(5y-7+4x)=0 [-8.62, 11.38, -8.44, 1.56]}