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

Answer 1

The correct option is A $7xy=6(x+y)$

$\frac{x}{3}+\frac{y}{4}=1$ and $\frac{x}{4}+\frac{y}{3}=1$

These can be simplified as
$4x+3y=12$ --(1)
$3x+4y=12$ --(2)

Multiply equation (1) by $3$ and (2) by $4,$
$12x+9y=36$
$12x+16y=48$

Eliminating $x,$ we get
$y=\frac{12}{7}$

Putting $y$ in equation (1), $x=\frac{12}{7}$

A variable line passes through $(\frac{12}{7},\frac{12}{7})$ and meets coordinate axis at $A$ and $B,$

Let the $y-\frac{12}{7}=m(x-\frac{12}{7})$

Point A is $\left(\frac{12}{7}\right(1-\frac{1}{m}),0)$

Point B is $(0,\frac{12}{7}(1-m\left)\right)$

midpoint is $\left(\frac{6}{7}\right(1-\frac{1}{m}),\frac{6}{7}(1-m\left)\right)$

$h=\frac{6}{7}(1-\frac{1}{m})$

$k=\frac{6}{7}(1-m)$

Eliminate $m,$ we get locus as
$7hk=6(h+k)$
$7xy=6(x+y)$