Question:
Find the value of $p$ so that the three lines $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$ may intersect at one point.
Solution:
The equations of the given lines are
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2x – y – 3 = 0 … (3)
On solving equations (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 line (2).
p (1) + 2 (–1) – 3 = 0
p – 2 – 3 = 0
p = 5
Thus, the required value of p is 5.
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