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
4x−3y+8=0
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Solution
The correct option is A4x+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=3−02+2=34 Slope of the required line =−43 Equation of the required line is, y−0=−43(x+2) ⇒4x+3y+8=0