Question

Find the value of $p$ so that the three lines $3x+y-2=0,px+2y-3=0$ and $2x-y-3=0$ may intersect at one point.

Answer 1

The equation of the given lines are
$3x+y-2=0$...(1)
$px+2y-3=0$....(2)
$2x-y-3=0$....(3)
On solving equation (1) and (3) we obtain
$x=1$ and $y=-1.$
Since these three lines may intersect at one point, the point of intersection of lines(1) and (3) will also satisfy lines(2)
$p\left(1\right)+2(-1)-3=0$
$p-2-3=0\Rightarrow p=5.$