Question

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

Answer 1

`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]}