A) \[x-3y=1\]
B) \[3x-2y=1\]
C) \[2x-3y=1\]
D) \[2x-y=1\]
Correct Answer: B
Solution :
The point of intersection of the lines is (1, 1). The equation of line parallel to \[2y-3x+2=0\] is \[2y-3x+k=0\]. It also passes through (1, 1), therefore\[AC=\sqrt{{{(4-0)}^{2}}+{{\left( 0-\frac{4}{3} \right)}^{2}}}=\frac{4\sqrt{10}}{3}\]. Hence the required equation is \[2y-3x+1=0\] or \[3x-2y=1\].You need to login to perform this action.
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