Question

A variable line is drawn through the intersection point of the lines $3x+4y-12=0$ and $x+2y-5=0$ meeting the coordinate axes at the points $A$ and $B$. Locus of mid point of segment $AB$ is

• A. $4x+3y=4xy$
• B. $3x+4y=3xy$
• C. $3x+4y=4xy$
• D. none

Answer 1

Answer: (C)

$3x+4y=4xy$

REF.Image

Integration of $(3x+4y-12=0\&x+2y=5)$

$x=5-2y$

$3x+4y=12$

$\Rightarrow 3(5-2y)+4y=12$

$\Rightarrow 15-6y+4y=12$

$\Rightarrow 2y=3\Rightarrow y=3/2$

$x=2$

Intersection pt. $\to (2,3/2)$

eqn. of line : $y=mx+c$

$\frac{3}{2}=2m+c\Rightarrow c=\frac{3}{2}-2m$

$y=mx+c\stackrel{x=0}{\to }y=c$

$\Rightarrow k=c$

$y=mx+c\stackrel{y=0}{\to }x=-c/m$

$\Rightarrow h=-c/m$

Mid pt. of $(h,0)\&(0,k)$ is $(h/2,k/2)$

$(\frac{h}{2},\frac{k}{2})-(\frac{-c}{2m},\frac{c}{2})=(\frac{2m-3/2}{2m},\frac{3/2-2m}{2})$

$=(1-\frac{3}{4m},\frac{3}{4}-m)=(h^{\prime },k^{\prime })$

$h^{\prime }=1-3/4m\Rightarrow \frac{3}{4m}=1-h^{\prime }\Rightarrow m=\frac{3}{4(1-h^{\prime })}$

$k^{\prime }=\frac{3}{4}-m\Rightarrow m=(\frac{3}{4}-k^{\prime })$

$m=\frac{3}{4(1-h^{\prime })}=\frac{3}{4}-k^{\prime }\Rightarrow \frac{3}{1-h^{\prime }}=3-4k^{\prime }$

$\Rightarrow 3=3(1-h^{\prime })-4k^{\prime }(1-h^{\prime })\Rightarrow 3h^{\prime }+4k^{\prime }=4h^{\prime }{k}^{\prime }$

$\Rightarrow 3x+4y=4xy$