A) \[3x+4y+5=0\]
B) \[3x+4y-10=0\]
C) \[3x+4y-5=0\]
D) \[3x+4y+6=0\]
Correct Answer: C
Solution :
The intersection point of lines \[x-2y=1\] and \[x+3y=2\] is \[\left( \frac{7}{5},\,\frac{1}{5} \right)\] and the slope of required line \[=-\frac{3}{4}\] \ Equation of required line is \[y-\frac{1}{5}=\frac{-3}{4}\left( x-\frac{7}{5} \right)\] Þ\[\frac{3x}{4}+y=\frac{21}{20}+\frac{1}{5}\]Þ\[3x+4y=5\]Þ\[3x+4y-5=0\].You need to login to perform this action.
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