Question

A straight line through the point $(1,1)$ meets the $X-$axis at $A$ and $Y-$axis at $B$. The locus of the midpoint of $AB$ is

• A. $2xy+x+y=0$
• B. $x+y–2xy=0$
• C. $x+y+2=0$
• D. $x+y-2=0$

Answer 1

The correct option is B

$x+y–2xy=0$

Explanation for the correct answer:

Step 1: Finding the intercept

Let $A(a,0)$ be the point on the x-axis and $B(0,b)$ be the point on the $Y-$axis.

Let a and b be the intercepts of $X-$axis and $Y-$axis.

The equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1...\left(1\right)$

The line passes through the point $P(1,1)$. So the equation $1$ becomes,

$$\begin{gathered} \Rightarrow \frac{1}{a}+\frac{1}{b}=1 \\ a+b=ab...\left(2\right) \end{gathered}$$

Step 2: Finding the midpoint

Let $Q(h,k)$ be the midpoint of the line $AB$ is,

$$\begin{gathered} \Rightarrow h=\frac{a+0}{2},k=\frac{0+b}{2} \\ h=\frac{a}{2},k=\frac{b}{2} \\ a=2h,b=2k \end{gathered}$$

Step 3: Finding the locus

Now substitute the value of $a$ and $b$ in equation $2$,

$$\begin{gathered} \Rightarrow 2h+2k=4hk \\ 2h+2h-4hk=0 \\ h+k-2hk=0 \end{gathered}$$

Thus the equation of the locus is $x+y-2xy=0$.

Hence, option (B) $\mathit{x}\mathbf{+}\mathit{y}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}\mathbf{=}\mathbf{0}$ is the correct answer.