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  • The equation of a line passing through the point of intersection of lines 2x-3y+4=0 , 3x+4y-5=0 and perpendicular to 6x-7y+3=0 then its equation is

The equation of a line passing through the point of intersection of lines 2x-3y+4=0 , 3x+4y-5=0 and perpendicular to 6x-7y+3=0 then its equation is

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Solution: The point of intersection of the lines \ 2x-3y+4=0 and \ 3x+4y-5=0 is \ (/-1/17,22/17)

The slope of required line =\ -7/6

Hence, equation of required line is \ y-22/7=-7/6(x+2/34)

\Rightarrow 119x+102y=125

Posted by

Deependra Verma

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