Question

The line $2x+3y=6,2x+3y=8$ cut the $X$-axis at $A,B$ respectively. $A$ line $L=0$ drawn through the point $(2,2)$ meets the $X$-axis at $C$ in such a way that abscissa of $A,B,C$ are in arithmetic progression. Then, the equation of the line $L$ is

• A. $2x+3y=10$
• B. $3x+2y=10$
• C. $2x-3y=10$
• D. $3x-2y=10$

Answer 1

The correct option is A $2x+3y=10$

Equation of line drawn through $(2,2)$ is given by
$y-2=m(x-2)$

$y=mx+2-2m$

at x-axis, $y=0,x=\frac{2(m-1)}{m}$ $c=(\frac{2(m-1)}{m},0)$

as given abscissae of $A,B,C$ are in A.P.

$\therefore 2\times 4=3+\frac{2(m-1)}{m}$ [abscissa = x- co ordinate]

$8=3+\frac{(m-1)}{m}2$ $\Rightarrow 5m=2m-2$
$3m=-2$

So, equation of line is $\Rightarrow m=\frac{-2}{3}$

$y=\frac{-2}{3}x+2+\frac{2\times 2}{3}$

$y=\frac{-2}{3}x+\frac{10}{3}$

$\Rightarrow 3y+2x=10.$