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Question

The area of the triangle formed by the lines, y = x, x + y = 2 and the line through p(h, k) and parallel to the x-axis is 4h2, then the point P lies on


A
2x - y + 1 = 0
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B
2x - y + 3 = 0
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C
x - y + 1 = 0
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D
x + 2y - 1 = 0
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Solution

The correct option is A 2x - y + 1 = 0
Given lines are y = x . . . (1)
x + y = 2 . . . (2)
and y = k . . . (3)
solving (1) & (2) A = (1, 1)
solving (2) & (3) B = (2 - k, k)
solving (1) & (3) C = (k, k)
given area of triangle = 4h2
12∣ ∣1112kk1kk1∣ ∣=4k2
(k1)2=4h2k1=±2h
(h,k)lies on 2x-y+1=0 (or) 2x+y-1=0

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