what is the equation of the line perpendicular to Ax + By = C through point (a,b)?

Question

what is the equation?  Then, after that equation, how do I find the point where that line and Ax + By = C intersect?

Answer 1

This is a standard form of a straight line. Write your equation in the slope - intercept form by subtracting the first term Ax from both sides of the equation and then dividng by B. You will obtain

  y = -(A/B)x + C/B                                        (1)

As you can see the slope "m"of this line is m = -(A/B). The slope "m₁" of the line perpendicular to (1) is defined as a reciprocal to "m" with negative sign: m₁ = -1/m. Thus we have for the equation of the perpendicular line:

        y = (B/A)x + c                                             (2)

where "c" is to be found. In order to find "c" we have to use the point (a,b). That means, if  x = a then y=b, or

       b = (B/A) a + c                                          (3)

Thus,  c = b - (B/A)a, and equation (2) can be rewritten 

      y = (B/A)x + b - (B/A) a = (B/A)(x-a) + b                 (4)

Because the point of intersection belongs to both lines, we can make equal y-s of two (1 and 4). Thus we have

         -(A/B) x + C/B = (B/A) x + b - (B/A)a                    (5)

Now solve (5) for x. You will obtain

        x = [(C/B) +(B/A)a -b]/[(B/A) +(A/B)]                     (6)

Now you have left tom plug (6) into (1) to find y. It gives you

         y = - (A/B)[(C/B) +(B/A)a - b]/[(B/A) + (A/B)] + C/B   (7)

Formulas (6) and (7) give solution of the problem in a very generic form.

Answer 2

Ax + By = C is perpendicular to Bx - Ay = D because {A, B} • {B, -A} = 0

So, the answer is Bx - Ay = (aB - bA), standard form.

Solve the system with determinants,

Ax + By = C ......(1)

Bx - Ay = (aB - bA) ......(2)

det = -(A^2+B^2)

x = [AC + B(aB - bA)]/(A^2+B^2)

y = -[A(aB - bA) + BC]/(A^2+B^2)