A) \[119x+102y+125=0\]
B) \[119x+102y=125\]
C) \[119x-102y=125\]
D) None of these
Correct Answer: B
Solution :
The point of intersection of lines \[2x-3y+4=0\] and \[3x+4y-5=0\] is \[\left( \frac{-2}{34},\,\frac{22}{17} \right)\] The slope of required line \[=\frac{-7}{6}\] \ Equation of required line is \[y-\frac{22}{17}=\frac{-7}{6}\left( x+\frac{2}{34} \right)\] Þ \[119x+102y=125\].You need to login to perform this action.
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