Browse Questions

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

$\begin {array} {1 1} (A)\;p=-5 & \quad (B)\;p=+5 \\ (C)\;p = \pm 5 & \quad (D)\;\text{ none of the above} \end {array}$

Toolbox:
• If three lines may intersect at one point, then the point of intersection of lines (1) and (3) will also satisfy line (2).
The equation of the given lines are
$\qquad 3x+y-2=0$---------(1)
$\qquad px+2y-3=0$---------(2)
$\qquad 2x-y-3=0$-----------(3)
Let us solve equation (1) and (3) we get,
$\qquad 3x+y-2=0$
$\qquad 2x-y-3=0$
$\qquad 5x-5=0 \Rightarrow x = 1$
$\qquad \therefore y = -1$
Hence the point of intersection of line (1) and line (3) is (1, -1)
Now substituting the value of $x$ and $y$ in equation (2), we get,
$\qquad p(1)+2(-1)-3=0$
$\qquad \Rightarrow p-5=0$
$\qquad \therefore p = 5$
Hence the required value is 5.