Question

If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$, meets the coordinate axes at $A$ and $B$, $(A \neq B)$, then the locus of the midpoint of $AB$ is :

• A. $6xy = 7(x +y)$
• B. $4(x+y)^2-28(x+y)+49=0$
• C. $7xy=6(x+y)$
• D. $14(x+y)^2-97(x+y)+168=0$

Answer 1

Answer: (C)

$L_{1} : 4x+3y-12=0$
$L_{2} : 3x+4y-12=0$
$L_{1}+\lambda L_{2}=0$
$\left(4x+3y-12\right)+\lambda\left(3x+4y-12\right)=0$
$x\left(4+3\lambda-12\right)+y\left(3+4\lambda\right)-12\left(1+\lambda\right)=0$
Point $A\left(\frac{12\left(1+\lambda\right)}{4+3\lambda},0\right)$
Point $B\left(0, \frac{12\left(1+\lambda\right)}{3+4\lambda}\right)$
mid point $\Rightarrow h=\frac{6\left(1+\lambda\right)}{4+3\lambda}......\left(i\right)$
$k=\frac{6\left(1+\lambda\right)}{3+4\lambda} ......\left(ii\right)$
Eliminate $\lambda$ from (i) and (ii) then
$6\left(h+k\right)=>hk$
$6\left(x+y\right)=>xy$