Question

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 ?

• A. \(x + 7y+2=0\)
• B. \(7x-y- 36 = 0\)
• C. \(7x-y+ 36 = 0\)
• D. \(x = y\)
• E. \(x-7y+2 = 0\)

Answer 1

The Correct Option is B

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  \(7x-y-36=0\)

Answer 2

Line AB: \(4x - 3y - 17 = 0\)

Line AC: \(3x + 4y - 19 = 0\)

The general formula for the bisectors of two lines \(ax + by + c = 0\) and \(px + qy + r = 0\) is given by:

\[\frac{ax + by + c}{\sqrt{a^2 + b^2}} = \pm \frac{px + qy + r}{\sqrt{p^2 + q^2}}\]

Applying this to our lines AB and AC:

\[frac{4x - 3y - 17}{\sqrt{4^2 + (-3)^2}} = \pm \frac{3x + 4y - 19}{\sqrt{3^2 + 4^2}}\]

Simplifying the square roots:

\[\frac{4x - 3y - 17}{5} = \pm \frac{3x + 4y - 19}{5}\]

Multiplying both sides by 5:

\[4x - 3y - 17 = \pm (3x + 4y - 19)\]

Now we have two cases:

Case 1: Positive Sign

\[4x - 3y - 17 = 3x + 4y - 19\]

\[x - 7y + 2 = 0\] (This matches option E)

Case 2: Negative Sign

\[4x - 3y - 17 = -(3x + 4y - 19)\]

\[4x - 3y - 17 = -3x - 4y + 19\]

Therefore, the equation of one of the bisectors is \(x - 7y + 2 = 0\), which corresponds to option E.