Find the equation of the perpendicular bisector of the line joining P(2, -3) to Q(-5, 1)

Find the equation of the perpendicular bisector of the line joining P(2, -3) to Q(-5, 1)

Answer Details

Given P(2, -3) and Q(-5, 1)

Midpoint = (2+(−5)2,−3+12) $(\frac{2+(-5)}{2},\frac{-3+1}{2})$

= (−32,−1) $(\frac{-3}{2},-1)$

Slope of the line PQ = 1−(−3)−5−2 $\frac{1-(-3)}{-5-2}$

= −47 $-\frac{4}{7}$

The slope of the perpendicular line to PQ = −1−47 $\frac{-1}{-\frac{4}{7}}$

= 74 $\frac{7}{4}$

The equation of the perpendicular line: y=74x+b $y=\frac{7}{4}x+b$

Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).

−1=(74)(−32)+b $-1=(\frac{7}{4})(\frac{-3}{2})+b$

−1+218=138=b $-1+\frac{21}{8}=\frac{13}{8}=b$

∴ $\therefore$ The equation of the perpendicular bisector of the line PQ is y=74x+138 $y=\frac{7}{4}x+\frac{13}{8}$

≡8y=14x+13⟹8y−14x−13=0