Question

Find the equation of the line joining the points of intersection of $2x+y=4$ with $x-y+1=0$ and $2x-y-1=0$ with $x+y-8=0$?

• A. $2x+3y+6=0$
• B. $3x+2y+12=0$
• C. $3x-2y+1=0$
• D. None of these

Answer 1

The correct option is C $3x-2y+1=0$

$2x+y=4$ ....$\left(1\right)$

$x-y=-1$ ....$\left(2\right)$

$2x-y=1$ ....$\left(3\right)$

$x+y=8$ ....$\left(4\right)$

Solve $\left(1\right)$,$\left(2\right)$ and $\left(3\right)$,$\left(4\right)$

$\Rightarrow$ $\left(1\right)$+$\left(2\right)$ $\Rightarrow$ $3x=3$

$\Rightarrow$ $x=1,y=2$

$\Rightarrow$ $\left(3\right)$+$\left(4\right)$ $\Rightarrow$ $3x=9$

$\Rightarrow$ $x=3,y=5$

Therefore, $(x_1,y_1)=(1,2)$

$(x_2,y_2)=(3,5)$

Equation of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is

$\Rightarrow \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$

Substituting these values, we get

$\Rightarrow \frac{x-1}{3-1}=\frac{y-2}{5-2}$

$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}$

$\Rightarrow 3(x-1)=2(y-2)$

$\Rightarrow 3x-3=2y-4$

$\Rightarrow 3x-2y=-4+3$

$\Rightarrow 3x-2y=-1$

$\Rightarrow 3x-2y+1=0$

Therefore, the equation of the required line is $\Rightarrow 3x-2y+1=0$