Question

A variable line, drawn through the point of intersection of the straight lines $\frac{x}{3}+\frac{y}{2}=1$ & $\frac{x}{2}+\frac{y}{3}=1$, meets the coordinate axes in A & B. The locus of the mid point of AB is the curve

• A. $6(x+y)=10xy$
• B. $(x+y)=10xy$
• C. $xy=6(x+y)$
• D. $2(x+y)=6xy$

Answer 1

Answer: (A)

$6(x+y)=10xy$

The equation of line through the point os intersection of $\frac{x}{3}+\frac{y}{2}-1=0$ and $\frac{x}{2}+\frac{y}{3}-1=0$ is of the form
$(\frac{x}{3}+\frac{y}{2}-1)+\lambda (\frac{x}{2}+\frac{y}{3}-1)=0$

Putting $y=0$ we get point $A$ as $\{\frac{6(1+\lambda )}{2+3\lambda },0\}$

Putting $x=0$ we get point $B$ as $\{0,\frac{6(1+\lambda )}{3+2\lambda }\}$

If $(h,k)$ be the mid-point of $AB$, then

$2h=\frac{6(1+\lambda )}{2+3\lambda }$ and $2k=\frac{6(1+\lambda )}{3+2\lambda }$

In order to find the locus of $(h,k)$ we have to eliminate the variable $\lambda$

Now $\frac{1}{2h}+\frac{1}{2k}=\frac{(2+3\lambda )+(3+2\lambda )}{6(1+\lambda )}=\frac{5(1+\lambda )}{6(1+\lambda )}$

$\Rightarrow \frac{h+k}{2hk}=\frac{5}{6}$

$\Rightarrow 6(h+k)=10hk$

Hence, the locus of mid-point $(h,k)$ is $6(x+y)=10xy$.