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 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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Solution

The correct option is B $4 x + 3 y + 8 = 0$
Given line
$a (3 x + 4 y + 6) + b (x + y + 2) = 0$

Here $L_{1} : 3 x + 4 y + 6 = 0$
$L_{2} : x + y + 2 = 0$

On solving $L_{1}, L_{2}$ we get
$3 x + 4 y = - 6$
$4 x + 4 y = - 8$
$x = - 2, y = 0$

Point of intersection of $L_{1}, L_{2}$ is $(- 2, 0)$

Now given point is $P (2, 3)$

Equation of line from $(- 2, 0)$ and $(2, 3)$
$y = \frac{3}{4} (x + 2)$

$4 y = 3 x + 6$

So the slope of above line is $\frac{3}{4}$

Line perpendicular to line $4 y = 3 x + 6$ passing through the common point will be at greatest distance

Hence Slope of required line be $- \frac{4}{3}$
$y = - \frac{4}{3} (x + 2)$

$3 y = - 4 x - 8$

$4 x + 3 y + 8 = 0$

Suggest Corrections

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