Question

The lines $2x+3y=6$, $2x+3y=8$ 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 the 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 $2x+3y=10$

Line $2x+3y=6$ meets $x-$ axis at $A(3,0)$
Line $2x+3y=8$ meets $x-$ axis at $B(4,0)$
Let $C$ be $(c,0)$
as given abscissa of $A,B,C$ are in A.P.
$\therefore 2\times 4=3+c[\text{abscissa}=x-\text{coordinate}]$
$\Rightarrow c=5$
So, equation of line passing through $(2,2)$ and $(5,0)$ is
$\frac{y-0}{2-0}=\frac{x-5}{2-5}$
$\Rightarrow 3y+2x=10$