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] : Given equations of straight lines, are \[4x-3y+7=0\]and \[3x-4y+14=0\] Here, \[{{a}_{1}}=4,{{b}_{1}}=-3\]and \[{{a}_{2}}=3,{{b}_{2}}=-4\] Now, \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=4\times 3+(-3)(-4)=24>0\] So, the bisector of the acute angle is given by \[\frac{4x-3y+7}{\sqrt{{{4}^{2}}+{{(-3)}^{2}}}}=-\frac{3x-4y+14}{\sqrt{{{3}^{2}}+{{(-4)}^{2}}}}\Rightarrow x-y+3=0\]
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