Given the family of lines, $(3 x + 4 y + 6) + λ (x + y + 2) = 0$ . The line of the family situated at the greatest distance from the point $P (2, 3)$ has equation

4 x + 3 y + 8 = 0

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5 x + 3 y + 10 = 0

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15 x + 8 y + 30 = 0

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5 x + 3 y + 6 = 0

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Solution

The correct option is A $4 x + 3 y + 8 = 0$
Let,
$L_{1} : 3 x + 4 y + 6 = 0$
$L_{2} : x + y + 2 = 0$
Line situated at greatest distance from $P (2, 3)$ is the line passing through point of intersection $(Q)$ of the given lines and perpendicular to $P Q .$
Point of intersection of Lines $L_{1}$ and $L_{2}$ is $Q (- 2, 0)$ .Let slope of farthest line be $m$ ,then we know that product of slopes of perpendicular lines is $- 1$
$m \times \frac{3 - 0}{2 - (- 2)} = - 1 ⟹ m = - \frac{4}{3}$ then
Equation of farthest line from given point is $\frac{y - 0}{x - (- 2)} = - \frac{4}{3} ⟹ 3 y + 4 x + 8 = 0$ .

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