Question

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 mid-point P of BC is

• A. x+y=1
• B. x+y=2
• C. x+y=2xy
• D. 2x+2y=1

Answer 1

The correct option is A x+y=1

Let $(h,k)$ be the mid point on the line $BC$.

Equation of $AB\to$

$(y-1)=m(x-1)$

Since given that $AB\perp AC$,

Therefore,

Equation of $AC\to$

$(y-1)=-\frac{1}{m}(x-1)$

Now, from fig.,

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

$\Rightarrow \frac{1}{m}=1-2h.....\left(1\right)$

Again from fig.,

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

$\frac{1}{m}=2k-1.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$1-2h=2k-1$

$\Rightarrow 2(h+k)=2$

$h+k=1$

Replacing $h$ and $k$ with $x$ and $y$ repectively, we hve

$x+y=1$

Hence the locus of mid-point of $BCC$ is $x+y=1$

Hence the correct answer is $\left(A\right)x+y=1$.