Given the family of lines $a (3 x + 4 y + 6) + b (x + y + 2) = 0.$ The line of the family situated at the greatest distance from the point $P (2, 3),$ has the 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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4 x - 3 y + 8 = 0

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Solution

The correct option is A $4 x + 3 y + 8 = 0$
Given the family of lines: $a (3 x + 4 y + 6) + b (x + y + 2) = 0$
So, the family of lines pass through the point of intersection of $3 x + 4 y + 6 = 0$ and $x + y + 2 = 0.$
$\Rightarrow$ Point of intersection $= (- 2, 0)$

The line situated at the greatest distance from the point $P (2, 3)$ is perpendicular to $A P .$
Slope of $A P = \frac{3 - 0}{2 + 2} = \frac{3}{4}$
Slope of the required line $= - \frac{4}{3}$
Equation of the required line is,
$y - 0 = - \frac{4}{3} (x + 2)$
$\Rightarrow 4 x + 3 y + 8 = 0$

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