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.

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Solution

The equation of the given lines are
$3 x + y - 2 = 0$ ...(1)
$p x + 2 y - 3 = 0$ ....(2)
$2 x - y - 3 = 0$ ....(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)
$p (1) + 2 (- 1) - 3 = 0$
$p - 2 - 3 = 0 \Rightarrow p = 5.$

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Similar questions

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

Find the value of k for which the lines $3 x + y = 2, k + 2 y = 3$ and $2 x - y = 3$ may intersect at a point.

k

2 x + y + k = 0

3 x^{2} - y^{2} = 3

3 x - y + 9 = 0

x + 2 y = 4

2 x + y - 4 = 0

x - 2 y + 3 = 0

k

3 x - k y - 1 = 0

4 x - y - 3 = 0

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