If each of the point $(a, 4), (- 2, b)$ lies on the line joining the points $(2, - 1), (5, - 3)$ , then the point $(a, b)$ lies on line

6 x + 6 y - 25 = 0

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x + 3 y + = 0

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2 x + 6 y + 1 = 0

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2 x + 3 y - 5 = 0

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Solution

The correct option is D $2 x + 6 y + 1 = 0$
Equation of the line passing through the point $(2, - 1)$ and $(5, - 3)$ is

$\frac{x - 2}{y + 1} = \frac{5 - 2}{- 3 + 1}$ $\Rightarrow \frac{x - 2}{y + 1} = \frac{3}{- 2}$

$\Rightarrow - 2 x + 4 = 3 y + 3$

$\Rightarrow 2 x + 3 y - 1 = 0.... (i)$

Also $(a, 4)$ and $(- 2, b)$ lies on the line $(i)$

$2 a + 3 (4) - 1 = 0$

$\Rightarrow 2 a + 12 - 1 = 0$

$\Rightarrow 2 a = - 11$

$\Rightarrow a = \frac{- 11}{2}$

$2 (- 2) + 3 b - 1 = 0$

$\Rightarrow - 4 + 3 b - 1 = 0$ $\Rightarrow 3 b = 5$

$\Rightarrow b = \frac{5}{3}$

$2 x + 6 y + 1 = 0 is the required line$ .

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