What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q showing work and final answer in its general form?

Answer 1

The perpendicular bisector of a line has a gradient (m') which is the negative reciprocal of the gradient (m) of the line, and passes through the mid-point of the line.

The equation of a line with gradient m through a point (x0, y0) has an equation of the form:

y - y0 = m(x - x0)

The gradient m of a line between two points (x0, y0) and (x1, y1) is given by:

m = change_in_y / change_in_x = (y1 - y0) / (x1 - x0)

Thus the line through (p, q) and (7p, 3q) has gradient:

m = (3q - q) / (7p - p) = 2q / 6p = q/3p

and the perpendicular bisector has gradient:

m' = -1 / m = -1 / (q/3p) = -3p/q

The midpoint of the line through (p, q) and (3p, 3q) is:

midpoint = ((p + 7p)/2), (q + 3q)/2) = (4p, 2q)

Thus the perpendicular bisector of the line between (p, q) and (7p, 3q) has equation:

y - 2q = -3p/q (x - 4p)

→ qy - 2q² = -3px + 12p²

→ qy + 3px = 12p² + 2q²

Additional Information:-

Final answer in its general form: 3px+qy-12p^2-2q^2 = 0