If the two sides AB and AC of a triangle are along \(4x-3y-17 = 0\) and \(3x+4y-19= 0\), then the equation of the bisector of the angle between AB and AC is ?
\(x + 7y+2=0\)
\(7x-y- 36 = 0\)
\(7x-y+ 36 = 0\)
\(x = y\)
\(x-7y+2 = 0\)
The Correct Option is B
Solution and Explanation
Given that
The two lines of a triangle ABC in which the line AB and AC passes through the \(4x-3y-17 = 0\) and \(3x+4y-19= 0\)
Then according to the question the Equation of Bisector of the angle can be found as follows
\(\dfrac{4x-3y-17}{√(4^{2}+(-3)^{2})} =± \dfrac{3x+4y-19}{√(3^{2}+4^{2})}\)
⇒\(4x-3y-17= ±(3x+4y-19)\)
taking Positive , the equation of bisector will be
\(x-7y-2=0\)
similarly taking the negative sign , the equation will be
\(7x-y-36=0\)
and as per the given option the right answer option \(7x-y-36=0\).. (Ans)
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