Question

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$, is

• A. $2x–3y+1=0$
• B. $6x–9y+11=0$
• C. $2x–3y+15=0$
• D. $3x–2y+1=0$

Answer 1

The correct option is A

$2x–3y+1=0$

Explanation of the correct answer.

Point of intersection of lines:

$x-3y+2=0$………$\left(1\right)$

$2x+5y-7=0$………$\left(2\right)$

Operation in equations $\left[2\times \left(1\right)\right]$$-$$\left(2\right)$

$\Rightarrow$$2x-6y+4-2x-5y+7=0$

$\Rightarrow$ $11y=11$

$\Rightarrow$ $y=1$

Substitute the value of $y$ in equation $\left(1\right)$,

$$\begin{gathered} x-3\left(1\right)+2=0 \\ \Rightarrow x=1 \end{gathered}$$

So the point of intersection of the lines is $\left(1,1\right)$.

Given: $3x+2y+5=0$

It can be written as, $y=-\frac{3}{2}x-\frac{5}{2}$…….$\left(3\right)$

Here is the slope of the line is $-\frac{3}{2}$.

Hence the slope of the line perpendicular to line $\left(3\right)$ is $\frac{2}{3}$.

Therefore, the equation of the required line is,

$$\begin{gathered} \left(y-1\right)=\frac{2}{3}\left(x-1\right) \\ \Rightarrow 3y-3=2x-2 \\ \Rightarrow 2x-3y+1=0 \end{gathered}$$

Hence, option $\left(\mathrm{A}\right)$ is the correct answer.