Question

The equation to the line which passes through the point of intersection of the two lines $2x+3y=1$ and $3x+2y+1=0$ and is normal to the line joining the points $(2,4),(4,7)$ is

• A. $3y+2x=1$
• B. $3x-2y=4$
• C. $2x+y+4=0$
• D. none of these

Answer 1

The correct option is A $3y+2x=1$

Equation of the line which passes through the point of intersection of the two lines $2x+3y=1$ and $3x+2y+1=0$ is
$(2x+3y-1)+\lambda (3x+2y+1)=0$
$\Rightarrow (2+3\lambda )x+(3+2\lambda )y+(\lambda -1)=0$ -----(1)
Let $A=(2,4)$ and $B=(4,7)$
Slope of $AB$ is $\frac{3}{2}$
but (1) is normal to $AB$.
$\Rightarrow$ Slope of (1) is $\frac{-2}{3}$
$\Rightarrow \frac{2+3\lambda }{3+2\lambda }=\frac{2}{3}$
$\Rightarrow \lambda =0$
$\therefore$ Required equation of line is $2x+3y=1$
Hence, option A.