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Q.

Given $A = (1, 1)$ and AB is any line through it cutting the x-axis in B. if AC is perpendicular to AB and meets the y-axis in C, then the equation of locus of midpoint P of BC is

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a

$x + y = 1$

b

$x + y = 2$

c

$x + y = 2 x y$

d

$2 x + 2 y = 1$

answer is A.

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Detailed Solution

Equation of line AB is $y - 1 = m (x - 1)$

Equation of line AC is $y - 1 = - \frac{1}{m} (x - 1)$

$2 h = 1 - \frac{1}{m}$

$2 k = 1 + \frac{1}{m}$

Eliminating m we have locus $x + y = 1$

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Given A=(1,1) and AB is any line through it cutting the x-axis in B. if AC is perpendicular to AB and meets the y-axis in C, then the equation of locus of midpoint P of BC is