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प्रश्न
Find the equation of the locus of the point P such that the line segment AB, joining the points A(1, −6) and B(4, −2), subtends a right angle at P
उत्तर
Given A(1, – 6) and B(4, – 2).
Let P(h, k) be a point such that the line segment AB subtends a right angle at P.
∴ ∆APB is a right-angled triangle.
AB2 = AP2 + BP2 ......(1)
AB2 = (4 – 1)2 + (– 2 + 6)2
AB2 = 32 + 42 = 9 + 16 = 25
AP2 = (h – 1)2 + (k + 6)2
BP2 = (h – 4)2 + (k + 2)2
(1) ⇒
25 = (h – 1)2 + (k + 6)2 + (h – 4)2 + (k + 2)2
25 = h2 – 2h + 1 + k2 + 12k + 36 + h2 – 8h + 16 + k2 + 4k + 4
25 = 2h2 + 2k2 – 10h + 16k + 57
2h2 + 2k2 – 10h + 16k + 57 – 25 = 0
2h2 + 2k2 – 10h + 16k + 32 = 0
h2 + k2 – 5h + 8k + 16 =0
The locus of P(h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x2 + y2 – 5x + 8y + 16 = 0
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