Determine a perpendicular bisector

Endpoint A

\(\displaystyle (1,2)\)

Endpoint B

\(\displaystyle (7,6)\)

Which equation is the perpendicular bisector of the segment?

Move X along the dashed vertical bisector. The two sloping segments join X to A and B. Compare their lengths and the squared distances below

What happens to the two distances as X moves?

Two conditions define the bisector

The line must pass through the midpoint and be perpendicular to the segment. Either condition on its own is insufficient

1. 1Midpoint Average corresponding endpoint coordinates \(\displaystyle M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right)\)
2. 2Perpendicular direction For an oblique segment, find its gradient and take the negative reciprocal \(\displaystyle \begin{gathered}m_{AB}=\frac{y_B-y_A}{x_B-x_A}\\m_{\perp}=-\frac1{m_{AB}}\end{gathered}\)
3. 3Use the midpoint Substitute the midpoint into the new line equation to find its intercept \(\displaystyle y_M=m_{\perp}x_M+c\)
4. 4Horizontal or vertical segment A horizontal segment has a vertical bisector. A vertical segment has a horizontal bisector \(\displaystyle x=x_M\quad\text{or}\quad y=y_M\)

Endpoint P

\(\displaystyle (-2,1)\)

Endpoint Q

\(\displaystyle (6,5)\)

Find the vertical-axis intercept of the perpendicular bisector

Find the midpoint, take the negative reciprocal of the segment gradient, then use the midpoint to find the intercept

Endpoint A

\(\displaystyle (-3,-4)\)

Endpoint B

\(\displaystyle (-3,2)\)

Which equation is the perpendicular bisector of this vertical segment?

Check the construction and write general form

For endpoints P and Q, combine the midpoint and perpendicular gradient before changing the equation’s form

1. 1Find the two ingredients Average the coordinates and calculate the gradient using matching differences \(\displaystyle \begin{aligned}M&=\left(\frac{-2+6}{2},\frac{1+5}{2}\right)=(2,3)\\m_{PQ}&=\frac{5-1}{6-(-2)}=\frac12\\m_{\perp}&=-2\end{aligned}\)
2. 2Use the midpoint Substitute the midpoint into the gradient-intercept equation \(\displaystyle \begin{aligned}3&=-2(2)+c\\c&=7\\y&=-2x+7\end{aligned}\)
3. 3General equation Move all terms to the left. For primitive general form, use integer coefficients with greatest common divisor one and the first nonzero coefficient positive \(\displaystyle \begin{gathered}2x+y-7=0\\(A,B,C)=(2,1,-7)\end{gathered}\)
4. 4Check the midpoint The completed equation contains the midpoint. The negative reciprocal already establishes its direction \(\displaystyle 2(2)+3-7=0\)

Endpoint P

\(\displaystyle (0,0)\)

Endpoint Q

\(\displaystyle (4,2)\)

Where is the first error in this attempted perpendicular bisector?