The lines $2 x+3 y=6,2 x+3 y=8$ cut the $X$-axis at $A$ and $B$, respectively. A line $L$ drawn through the…

The lines $2 x+3 y=6,2 x+3 y=8$ cut the $X$-axis at $A$ and $B$, respectively. A line $L$ drawn through the point $(2,2)$ meets the $X$-axis as $C$ in such a way that abscissae of $A, B$ and $C$ are in arithmetic progression. Then, the equation of the line $L$ is
  1. $2 x+3 y=10$
  2. $8 x+2 y=10$
  3. $2 x-3 y=10$
  4. $8 x-2 y=10$

Solution

Given lines are
$2 x+3 y=6$ and $2 x+3 y=8$ These lines meet the $X$-axis at $A(3,0)$ and $B(4,0)$. The line $L$ passes through $(2,2)$ cuts $X$-axis at $C$, such that $x$-coordinate of $A, B, C$ are in ap i.e. $3,4, x_1$ $\begin{array}{ll}\Rightarrow & 2.4=3+x_1 \\ \Rightarrow & x_1=8-3=5\end{array}$ To coordinates of $C$ are $(5,0)$. A line $L$ passing through the points $(2,2)$ and $(5,0)$, is $\begin{aligned} & y-0 & =\frac{0-2}{5-2}(x-5) \\ \Rightarrow & y & =\frac{-2}{3}(x-5) \\ \Rightarrow & 3 y & =-2 x+10 \\ \Rightarrow & 2 x+3 y & =10\end{aligned}$

Asked in: AP EAMCET 2001

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