Question

Find the equation of the straight line passing through the point of intersection of the lines $3x-y+9=0$ and $x+2y=4$ and the point of intersection of the lines $2x+y-4=0$ and $x-2y+3=0$.

• A. $x+3y-7=0$
• B. $x+3y+7=0$
• C. $x+3y-9=0$
• D. None of these

Answer 1

The correct option is A $x+3y-7=0$

Consider the given equation of the line.

$3x-y+9=0$ $...........\left(1\right)$

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

By eqn $3\times \left(2\right)-\left(1\right)$

We get, $y=3$, putting in eqn $\left(1\right)$

we get $x=-2$

Point $\left(P\right)$ of intersection of above both lines is

$P=(-2,3)$

Consider the given other equations of the line.

$2x+y-4=0$ $...........\left(3\right)$

$x-2y+3=0$ $...........\left(4\right)$

By eqn $2\times \left(3\right)+\left(4\right)$

We get, $x=1$, putting in eqn $\left(4\right)$

we get $y=2$

Point $\left(Q\right)$ of intersection of above both lines is

$Q=(1,2)$

So, the equation of line passes through intersection points $P$ and $Q$

$y-3=\frac{2-3}{1+2}(x+2)$

$y-3=\frac{-1}{3}(x+2)$

$3y-9=-x-2$

$x+3y-7=0$

Hence, this is the answer.