Question

Let A,B, C be angles of the triangle with vertex $A\equiv \left( 4,-1 \right)$ and internal angular bisectors of angles B and C be $x-1=0$ and $x-y-1=0$ respectively.
Slope of BC is
(a)$\dfrac{1}{2}$
(b)2
(c)3
(d)12

Answer 1

Hint: First find the points A, B, C by intersection of sides given in the question. So, given an internal angular bisector, the bisector bisects the angle into 2 equal halves hence, it divides the side opposite into 2 parts with ratio of corresponding sides. Equation of angular bisector of two equations of lines $ax+by+c=0$ and $dx+ey+f=0$ is given by
$\dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\pm \dfrac{dx+ey+f}{\sqrt{{{d}^{2}}+{{e}^{2}}}}$
These 2 lines will become an equation of bisectors.

Complete step-by-step answer:
If two lines form an angle, then they have 2 angle bisectors because between 2 lines there are 2 angles possible which are acute and obtuse. So, the 2 lines bisecting 2 angles between lines $ax+by+c=0$ , $dx+ey+f=0$ are given by
$\dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\pm \dfrac{dx+ey+f}{\sqrt{{{d}^{2}}+{{e}^{2}}}}$
So, the symbol plus or minus denotes 2 equations of bisectors. Out of both any of them may be acute and may be obtuse. If one is acute, the other is obtuse and vice-versa.
By general knowledge of geometry, we can say the image of vertex on an angular bisector always lies on the side opposite. So, we have A and 2 bisectors.
By general knowledge of geometry, image of point $\left( a,b \right)$ with respect to $px+qy+r=0$ is $\left( h,k \right)$ then the formula will be
$\dfrac{h-a}{p}=\dfrac{k-b}{q}=\dfrac{-2\left( pa+qb+r \right)}{{{p}^{2}}+{{q}^{2}}}$
Image of point $\left( 4,-1 \right)$ in bisectors of B lies on BC.
Image of point $\left( 4,-1 \right)$ in bisectors of C lies on BC.
So, first we take $\left( 4,-1 \right)$ in the equation $x-y-1=0$
$\dfrac{x-4}{1}=\dfrac{y+1}{-1}=-2\dfrac{\left( 4\left( -1-1 \right) \right)}{2}$
By equating x-term and simplifying it, we get that the:
$x=4-2\dfrac{\left( 4+1-1 \right)}{2}=4-4-1+1=0$
By equating y-term and simplifying for y, we get that
$y=-1+2\dfrac{\left( 4+1-1 \right)}{2}=3$
By above equations we get the coordinates as $\left( 0,3 \right)$
Let this point be represented by P.
Next, we take $\left( 4,-1 \right)$ in the line $x-1=0$ .
By using image formula mentioned above, we get:
$\dfrac{x-4}{1}=\dfrac{y+1}{0}=\dfrac{-2\left( 4-1 \right)}{1}=-6$
By equating x-term and simplifying for x, we get that:
$x=4-6=-2$
By equating y-term and simplifying for y, we get that
$y=-1+0=-1$
By above equation, we get the co-ordinates of the point. Let it be named as Q. $Q=\left( -2,1 \right)$
By condition, we know P, Q lies in BC.
So, we can say that slopes of PQ and BC are equal.
Slope of line passing through $\left( a,b \right)$ and $\left( c,d \right)$ will be
Slope $=\dfrac{d-b}{c-a}$
By substituting the points, we get slope as:
Slope $=\dfrac{3+1}{0+2}=\dfrac{4}{2}=2$
By simplifying we get slope as 2.
Therefore, the slope of BC is 2.
Option (b) is correct.

Note: (1) Don’t forget the constant ‘ $-2$ ’ in the formula of image. Generally, students tend to forget that and solve unknowingly.
(2) Slope is always the ratio of some corresponding co-ordinates in numerator and denominator. If we take 2nd minus 1st in numerator take the same in denominator or else we will see an extra “ $-$ “ sign.