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 $B$, $(A\ne B)$, 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
$6\left(h+k\right)=>hk$
$6\left(x+y\right)=>xy$