# 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

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 $\\ y-22/7=-7/6(x+2/34)$

$\\\Rightarrow 119x+102y=125$

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