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)

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

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

= \((\frac{-3}{2}, -1)\)

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

= \(-\frac{4}{7}\)

The slope of the perpendicular line to PQ = \(\frac{-1}{-\frac{4}{7}}\)

= \(\frac{7}{4}\)

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

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

\(-1 = (\frac{7}{4})(\frac{-3}{2}) + b\)

\(-1 + \frac{21}{8} = \frac{13}{8} = b\)

\(\therefore\) The equation of the perpendicular bisector of the line PQ is \(y = \frac{7}{4}x + \frac{13}{8}\)

\(\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0\)

There is an explanation video available below.

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