Question

The value of $p$ if the lines $x+p=0,y-2=0$ and $3x+2y+5=0$ are concurrent is

Answer 1

The lines are
$x+p=0\cdots \left(1\right)y-2=0\cdots \left(2\right)3x+2y+5=0\cdots \left(3\right)$

Solving $\left(1\right)$ and $\left(2\right),$ we get their point of intersection i.e. $(-p,2)$.
$\because$ The three lines are concurrent, hence the point $(-p,2)$ must satisfy the equation $\left(3\right).$
$\therefore 3(-p)+2\left(2\right)+5=0$
$\Rightarrow p=3$