Equation of a line passing through the intersection of

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Solution

A) $L_{1} : 2 x + 3 y - 4 = 0$ and $L_{2} : 3 x + 4 y - 2 = 0$
point of intersection of $L_{1} :$ & $L_{2} :$ is $P (- 10, 8)$
Therefore, P lies on $x + y + 2 = 0$
B) $L_{1} : x + y + 2 = 0$ and $L_{2} : 7 x - 5 y + 6 = 0$
point of intersection of $L_{1} :$ & $L_{2} :$ is $P (\frac{- 4}{3}, \frac{- 2}{3})$
Therefore, P lies on $x - 2 y = 0$
C) $L_{1} : x - 2 y = 0$ and $L_{2} : 3 x + y - 4 = 0$
point of intersection of $L_{1} :$ & $L_{2} :$ is $P (\frac{8}{7}, \frac{4}{7})$
Therefore, P lies on $2 x + 3 y - 4 = 0$
D) $L_{1} : x + y + 1 = 0$ and $L_{2} : x - y - 1 = 0$
point of intersection of $L_{1} :$ & $L_{2} :$ is $P (0, - 1)$
Therefore, P lies on $y + 1 = 0$

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Similar questions

2 x + y - 3 = 0

3 x - y - 2 = 0

2 x + y = 1

3 x + 2 y = 5

\to r ._{1} = q_{1}

\to r ._{2} = q_{2}

x^{2} + x y - 2 y^{2} - 4 x + 7 y - 5 = 0

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