Given A ≡(1,1) and A B is any line through it cutting the x-axis at B. If A C is perpendicular to A B and meets the y-axis in C, then the equation of the locus of midpoint P of B C is

Question

Given A ≡(1,1) and A B is any line through it cutting the x-axis at B. If A C is perpendicular to A B and meets the y-axis in C, then the equation of the locus of midpoint P of B C is (A) x+y=1 (B) x+y=2 (C) x+y=2 x y (D) 2 x+2 y=1

Tardigrade Admin · Accepted Answer

The equation of line A B is y-1=m(x-1). Therefore, the equation of line A C is y-1=-(1/m)(x-1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2e274894a1dbfc2a3ab0398760b7bd1c-.png" /> 2 h=1-(1/m) 2 k=1+(1/m) Eliminating m, we have locus x+y=1.

Q. Given $A \equiv (1, 1)$ and $A B$ is any line through it cutting the $x$ -axis at $B$ . If $A C$ is perpendicular to $A B$ and meets the $y$ -axis in $C$ , then the equation of the locus of midpoint $P$ of $BC$ is

$x + y = 1$

$x + y = 2$

$x + y = 2 x y$

$2 x + 2 y = 1$

The equation of line $A B$ is $y - 1 = m (x - 1)$ .
Therefore, the equation of line $A C$ 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 $x + y = 1$ .

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

x+y=1

x+y=2

x+y=2xy

2x+2y=1