Question

A variable line drawn through the point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1$ meets the coordinate axes in $A$ and $B$. Then the locus of the mid point of $AB$ is

• A. $2xy(a+b)=ab(x+y)$
• B. $xy(a+b)=ab(x-y)$
• C. $xy(a+b)=ab(x+y)$
• D. $xy(a+b)=2ab(x+y)$

Answer 1

The correct option is A $2xy(a+b)=ab(x+y)$

Given lines are $\frac{x}{a}+\frac{y}{b}=1\cdots \left(1\right),\frac{x}{b}+\frac{y}{a}=1\cdots \left(2\right)$

$\left(1\right)-\left(2\right)⟹x(\frac{1}{a}-\frac{1}{b})+y(\frac{1}{b}-\frac{1}{a})=0⟹x+y=0⟹x=y$

Substituting $x=-y$ in $\left(1\right)$ gives $\frac{y}{a}+\frac{y}{b}=1⟹y=\frac{ab}{a+b}$

So $(x,y)=(\frac{ab}{a+b},\frac{ab}{a+b})$

Let the point on coordinate axes be $A(h,0)$ and $B(0,k)$

So the mid point of $AB$ be $(\frac{h}{2},\frac{k}{2})$

So the variable line will be $\frac{x}{h}+\frac{y}{k}=1$

The point $(\frac{ab}{a+b},\frac{ab}{a+b})$ lies on the variable line

$\frac{ab}{h(a+b)}+\frac{ab}{k(a+b)}=1⟹\frac{\frac{ab}{2}}{\frac{h}{2}(a+b)}+\frac{\frac{ab}{2}}{\frac{k}{2}(a+b)}=1$

So the locus of mid point of $AB$ is $\frac{ab}{2x(a+b)}+\frac{ab}{2y(a+b)}=1⟹2xy(a+b)=ab(x+y)$