# Thread: Equation of perpendicular bisector

1. ## Equation of perpendicular bisector

Next Question .... Sorry :P

The line with equation x+3y=12 meets the x and y axes at the points A and B respectively. Find the equation of the perpendicular bisector of AB

2. This requires a couple of steps. The words look all googly and weird but its pretty basic:

You have the equation for a line. You are told to find the bisector (the midpoint) of the line formed by the points A and B, which are your X and Y intercepts. Therefore:

$x+3y=12 \Longrightarrow x-12=3y \Longrightarrow \frac{12-x}{3}=y \Longrightarrow 4-\frac{1}{3}x=y$

To find the X-int, we set Y=0 and solve:

$0=4-\frac{1}{3}x \Longrightarrow \frac{1}{3}x=4 \Longrightarrow x=12$

Alternatively, we could have used out original equation.

To find the Y=int, we set X=0 and solve. This one is straight forward. Y=4.

Now we have the co-ordinates of point A and B: (12,0) and (0,4) respectively. Our question asks us to find the equation of the perp-bisector. Now for a line to be perpendicular to another line, it must have the a slope equal to the negative reciprocal of the line it is to be perpendicular to. The slope of our original line is simple to see: $-\frac{1}{3}$. Therefore the negative reciprocal is 3. Thus the slope of our perpendicular line is 3.

But we aren't done. We have a very specific point we need to use in order to make sure this line bisects the X and Y intercepts of our original equation. Thus, we need to use the midpoint formula $\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$, to find the midpoint between (12,0) and (0,4):

$\frac{12+0}{2},\frac{0+4}{2} \Longrightarrow (6,2)$

Now, we have a point (6,2), a slope of 3 and a form called the point-slope form of an equation:

$y-2=3(x-12) \Longrightarrow y=3x-36+2 \Longrightarrow y=3x-34$

3. Originally Posted by Math-DumbassX
Next Question .... Sorry :P

The line with equation x+3y=12 meets the x and y axes at the points A and B respectively. Find the equation of the perpendicular bisector of AB