Question

Find a general form of an equation for the perpendicular bisector of the segment AB.
A(4, −2), B(−1, 7)
I keep getting 5/9x+y=10/3

Answer 1

A(4, −2), B(−1, 7)
slope(AB) = \(\large \frac{7--2}{-1-4}=-\frac{9}{5} \)
so the slope you want is 5/9.

the midpoint is \(\large (\frac{4+-1}{2}, \frac{-2+-7}{2}) \rightarrow (\frac{3}{2}, \frac{-9}{2}) \)

Answer 2

since you have a point and a slope, use point-slope form:
\(\large y-(-\frac{9}{2})=(\frac{5}{9})(x-\frac{3}{2}) \)
\(\large y+\frac{9}{2}=\frac{5}{9}(x-\frac{3}{2}) \)
\(\large y+\frac{9}{2}=\frac{5}{9}x-\frac{15}{18} \)
\(\large y+\frac{9}{2}=\frac{5}{9}x-\frac{5}{6} \)
\(\large 18[y+\frac{9}{2}=\frac{5}{9}x-\frac{5}{6}] \)
\(\large 18y+18 \cdot \frac{9}{2}=18 \cdot \frac{5}{9}x-18 \cdot \frac{5}{6} \)
\(\large 18y+81=10x-15 \)

now put in general form....