Equation of Perpendicular Lines

Question

1. Perpendicular to 3y – 2x + 11 = 0 going through (2,–2) 
Answer to this question = 2y + 3x - 2 =0

 

2. Perpendicular to 2x – 3y = 4x + 2y – 11 going through (2,10)

Answer to this question = 5x – 2y + 10 = 0

 

 

I know how to do these in y=mx+c form but how to find the answers to these questions in Ax + By +c = 0 form(A, B and C can be any number.)

Answer 1

You have just asked one of my favorite questions on WyzAnt's Answers Forum.  So, here goes --

 

First, you must know that perpendicular lines have negative reciprocal slopes.  You know that from y=mx+b and y=(-1/m)x + d.

 

Now, do this ONCE AND FOREVER: convert  Ax + By = C  to  y = (-A/B)x + (C/B)  and remember it.  The slope of the line is (-A/B).

 

Of course, that means that the slope of all perpendicular lines is (B/A).  So, using the original values of A and B, the equation of all perpendicular lines will have the negative of swapped coefficients and a different constant!  We need a point to find the constant.

1. Perpendicular to 3y – 2x + 11 = 0 going through (2,–2) 

   Perpendicular line is   +2y + 3x + D =0  and we need point (2,-2) for  2(-2)+3(2) +D=0  so D=-2

       (note that this problem writes D on left side of equation)

 
Answer to this question = 2y + 3x - 2 =0
 
2. Perpendicular to 2x – 3y = 4x + 2y – 11 going through (2,10)

      This is   2x + 5y -11 = 0

    Swap (5) with (2) and apply (-); Use (2,10) to find D.

Answer to this question = 5x – 2y + 10 = 0

Answer 2

When trying to solve problems like this, the first step is to restate the equation in the point intercept form, i.e. y-mx+b

 

1. 3y-2x+11 = 0

y = 2/3 x - 11/3

 

2. 2x-3y=4x+2y-11

5y = -2x+11

y = -2/5 x + 11/5

 

From this form, we can easily see the slope

1. 2/3

2 -2/5

 

The slope of the line perpendicular to the given line will be the negative inverse of the slope

1. -3/2

2. 5/2

 

Now for the perpendicular line through a given point.  It's best to use the point slope form which looks like

y - a(x-h)² + k

 

1. y=(-3/2)(x-2)^2-2

2. y=(5/2)(x-2)²+10

 

Finally, expand these equations to obtain the answer in the required form.

1. y=(-3/2)(x²-4x+4)-2 = -3/2 x²-6x-5

-3/2 x² - 6x - y - 5 = 0

-3x² -12x -2y - 10 = 0

 

and 

2. y = (5/2)(x²-4x+4)+10

y = 5/2 x²-10x+20

5/2 x² - 10x - y + 20 =0

5x² - 20x - 2y + 40 = 0