Solveeit Logo
Login

Question

Question: Let A,B, C be angles of the triangle with vertex \(A\equiv \left( 4,-1 \right)\) and internal angula...

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

Explanation

Solution

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=0ax+by+c=0 and dx+ey+f=0dx+ey+f=0 is given by
ax+by+ca2+b2=±dx+ey+fd2+e2\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=0ax+by+c=0 , dx+ey+f=0dx+ey+f=0 are given by
ax+by+ca2+b2=±dx+ey+fd2+e2\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 (a,b)\left( a,b \right) with respect to px+qy+r=0px+qy+r=0 is (h,k)\left( h,k \right) then the formula will be
h−ap=k−bq=−2(pa+qb+r)p2+q2\dfrac{h-a}{p}=\dfrac{k-b}{q}=\dfrac{-2\left( pa+qb+r \right)}{{{p}^{2}}+{{q}^{2}}}
Image of point (4,−1)\left( 4,-1 \right) in bisectors of B lies on BC.
Image of point (4,−1)\left( 4,-1 \right) in bisectors of C lies on BC.
So, first we take (4,−1)\left( 4,-1 \right) in the equation x−y−1=0x-y-1=0
x−41=y+1−1=−2(4(−1−1))2\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(4+1−1)2=4−4−1+1=0x=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(4+1−1)2=3y=-1+2\dfrac{\left( 4+1-1 \right)}{2}=3
By above equations we get the coordinates as (0,3)\left( 0,3 \right)
Let this point be represented by P.
Next, we take (4,−1)\left( 4,-1 \right) in the line x−1=0x-1=0 .
By using image formula mentioned above, we get:
x−41=y+10=−2(4−1)1=−6\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=−2x=4-6=-2
By equating y-term and simplifying for y, we get that
y=−1+0=−1y=-1+0=-1
By above equation, we get the co-ordinates of the point. Let it be named as Q. Q=(−2,1)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 (a,b)\left( a,b \right) and (c,d)\left( c,d \right) will be
Slope =d−bc−a=\dfrac{d-b}{c-a}
By substituting the points, we get slope as:
Slope =3+10+2=42=2=\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 ‘ −2-2 ’ 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.