Question

Let x + y = 2 and 2x + y = 5 are internal angle bisectors of angles B and C of \u0394ABC respectively. If A(3,4), then which of the following options is/are correct?

• A. Image of A about internal bisector of angle B is (-2,-1).
• B. If r denotes inradius of \u0394ABC, then 2r = 3\sqrt{10}
• C. Equation of BC is y = 3x + 5
• D. Image of A about internal bisector of angle C is (-1,-2).

Answer 1

Answer: (A), (C)

Image of A about internal bisector of angle B is (-2,-1); Equation of BC is y = 3x + 5

Answer 2

Step 1. Intersection of the two bisectors gives the in‑center
Solve

$$\begin{cases} x+y=2,\\ 2x+y=5, \end{cases}$$

Subtracting gives $x=3$, then $y=-1$. So $I=(3,-1)$.

Step 2. Reflection of $A(3,4)$ about line $x+y-2=0$
For line $ax+by+c=0$ and point $(x_1,y_1)$, the reflection is

$$d=\frac{ax_1+by_1+c}{a^2+b^2},\quad (x',y')=\bigl(x_1-2ad,\;y_1-2bd\bigr).$$

Here $a=1,b=1,c=-2$:

$$d=\frac{3+4-2}{2}=2.5,\quad (x',y')=(3-5,\;4-5)=(-2,\,-1).$$

Thus (A) is true.

Step 3. Reflection of $A$ about $2x+y-5=0$
Here $a=2,b=1,c=-5$:

$$d=\frac{6+4-5}{5}=1,\quad (x',y')=(3-4,\;4-2)=(-1,\;2).$$

This is not $(-1,-2)$, so (D) is false.

Step 4. Find $B,C$ by reflections

$$B=\text{reflection of }A\text{ across }2x+y-5=0 =(-1,2),\quad C=\text{reflection of }A\text{ across }x+y-2=0 =(-2,-1).$$

Step 5. Equation of $BC$
Through $B(-1,2)$ and $C(-2,-1)$, slope $m=\frac{2-(-1)}{-1-(-2)}=3$.

$$y-2=3(x+1)\;\Longrightarrow\;y=3x+5.$$

So (C) is true.

Step 6. Inradius check
Side‑lengths:
$BC=\sqrt{10},\;CA=5\sqrt2,\;AB=2\sqrt5$.
Semiperimeter $s=\tfrac{\sqrt{10}+5\sqrt2+2\sqrt5}{2}$.
Area by coordinates = 5.

$$r=\frac{\Delta}{s}=\frac{5}{s}\approx0.68\quad\Longrightarrow\quad2r\approx1.36\neq3\sqrt{10}.$$

So (B) is false.