Question

The variable straight line passing through the point of intersection of the lines $x+2y=1$ and $2x-y=1$ meet the co-ordinate axes in $A$ and $B$. the locus of the mid point of $AB$ is

• A. $10xy=x-3y$
• B. $10xy=x+3y$
• C. $xy=x+3y$
• D. None of these

Answer 1

Answer: (B)

$10xy=x+3y$

Equation of any line passes through the intersection of $x+2y-1=0and2x-y-1=0$ is $(x+2y-1)+\lambda (2x-y-1)=0$ $\lambda$ is a variable
$\Rightarrow x(2\lambda +1)+y(2-\lambda )-(\lambda +1)=0$
which meet the co - ordinate axes at
$A(\frac{\lambda +1}{2\lambda +1},0)andB(0,\frac{\lambda +1}{2-\lambda })$
Let $P(h,k)$ be the mid point of $AB$
$\Rightarrow h=\frac{\lambda +1}{2(2\lambda +1)},k=\frac{\lambda +1}{2(2-\lambda )}$
$\Rightarrow \frac{2\lambda +1}{\lambda +1}=\frac{1}{2h},\frac{(2-\lambda )}{\lambda +1}=\frac{1}{2k}$
$\Rightarrow \frac{1}{2h}+\frac{1}{2k}=\frac{\lambda +3}{\lambda +1}$
$\Rightarrow \frac{k+h}{2hk}=1+\frac{2}{\lambda +1}\Rightarrow \frac{2}{\lambda +1}=\frac{k+h-2hk}{2hk}$
$\Rightarrow \lambda +1=\frac{4hk}{k+h-2hk}\Rightarrow \lambda =\frac{6hk-k-h}{h+k-2hk}$
Now putting the value of $\lambda$ in $h=\frac{\lambda +1}{2(2\lambda +1)}$
$\Rightarrow h=\frac{2hk}{10hk-k-h}$
$\Rightarrow 10hk=h+3k$
$\therefore$ locus of $P(h,k)$ is $10xy=x+3y$
Hence, option B is the correct answer.