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Question

Given the family of lines a(3x+4y+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

A
4x+3y+8=0
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B
5x+3y+10=0
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C
15x+8y+30=0
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D
4x3y+8=0
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Solution

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


The line situated at the greatest distance from the point P(2,3) is perpendicular to AP.
Slope of AP=302+2=34
Slope of the required line =43
Equation of the required line is,
y0=43(x+2)
4x+3y+8=0

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