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Question
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)
Solution
Let P(x, y) be any point on the required locus.
Given, A(1, 6) and B(3, 5),
∠APB = 90°
∴ ΔAPB is a right angled triangle.
By Pythagoras theorem,
AP2 + PB2 = AB2
∴ [(x – 1)2 + (y – 6)2] + [(x – 3)2 + (y – 5)2] = (1 – 3)2 + (6 – 5)2
∴ x2 – 2x + 1 + y2 – 12y + 36 + x2 – 6x + 9 + y2 – 10y + 25 = 4 + 1
∴ 2x2 + 2y2 – 8x – 22y + 66 = 0
∴ x2 + y2 – 4x – 11y + 33 = 0
∴ The required equation of locus is
x2 + y2 – 4x – 11y + 33 = 0.
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