How do you write the standard form of a line given (-6,7) and perpendicular to 3x+4y-12=0?

1 Answer
Jul 25, 2017

4x-3y=-45

Explanation:

Since the equation must be expressed in standard form, that is, Ax+By=C, we can find the perpendicular line by switching the coefficients of x and y and changing the operation sign.

But first let's rewrite the given equation in Ax+By=C form.

3x+4y-12=0

3x+4y=12

To find the perpendicular line, switch the coefficients of x and y and change the operation sign. By doing so we get:

4x-3y=C; where C is any constant but we are given the point (-6,7) so we can substitute these values for x and y to find C.

If we let (-6,7)->(x,y) then,

4(-6)-3(7)=C

-24-21=C

-45=C

So we found our constant. We can now substitute this value for C for the equation of the perpendicular line. In doing so, we get our final equation:

4x-3y=-45larr And this is the equation of the perpendicular line