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.

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Solution

The equation of lines are

3x + y - 2 = 0, px + 2y - 3 = 0 and 2x -y - 3 = 0,

We know that three lines are concurrent if

$a_{3} (b_{1} c_{2} - b_{2} c_{1}) + b_{3} (c_{1} a_{2} - c_{2} a_{1}) + c_{3} (a_{1} b_{2} - a_{2} b_{1}) = 0$

$∴ 2 [1 \times (- 3) - 2 \times (- 2)] + (- 1) [- 2 \times p - (- 3) \times 3] + (- 3) [3 \times 2 - p \times 1] = 0$

$\Rightarrow 2 [- 3 + 4] - 1 [- 2 p + 9] - 3 [6 - p] = 0$

$\Rightarrow$ 2 + 2p - 9 - 18 + 3p = 0

$\Rightarrow$ 5p - 25 = 0

$\Rightarrow$ p = 5

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Similar questions

p

3 x + y - 2 = 0, p x + 2 y - 3 = 0

2 x - y - 3 = 0

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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