A) form a triangle
B) are only concurrent
C) are concurrent with one line bisecting the angle between the other two
D) None of the above
Correct Answer: C
Solution :
Equations of angle bisectors of lines \[{{\alpha }_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{\alpha }_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]are \[\frac{{{\alpha }_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{\alpha _{1}^{2}+b_{1}^{2}}}=\pm \frac{{{\alpha }_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{\alpha _{2}^{2}+b_{2}^{2}}}\]. For given two lines \[24x+7y-20=0\] and\[4x-3y-2=0\], the angle bisectors are given by \[\frac{24x+7y-20}{25}=\pm \frac{4x-3y-2}{5}\] Taking positive sign, we get \[2x+11y-5=0\] Therefore, the given three lines are concurrent with one line bisecting the angle between the other two.
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