A) \[x+y+3=0\]
B) \[x-y-3=0\]
C) \[x-y+3=0\]
D) \[3x+y-7=0\]
Correct Answer: C
Solution :
[c] If a point is equidistant from the two intersecting lines, then the locus of this point is the angle bisector of those lines. Now, let (h, k) be the point which is equidistant from the lines \[4x-3y+7=0\] and \[3x-4y+14=0\] Then \[\frac{4h-3k+7}{\sqrt{{{4}^{2}}+{{(-3)}^{2}}}}=\pm \frac{3h-4k+14}{\sqrt{{{3}^{2}}+{{(-4)}^{2}}}}\] \[\Rightarrow 4h-3k+7=\pm (3h-4k+14)\] \[\Rightarrow h+k-7=0\] and \[7h-7k+21=0\] Hence locus of \[(h,k)\] is \[x+y-7=0\] and \[x-y+3=0\]
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