Question

Consider a triangle $ABC$ whose vertices are $B (-1,3)$ and $C (3,6)$. Also the equation of internal angle bisector of $\angle BAC$ is $3 x + y =5$.
The area of triangle formed by joining midpoints of sides $BC , CA$ and $AB$ of triangle $ABC$ is equal to

• A. $\frac{3}{4}$
• B. 5
• C. $\frac{5}{4}$
• D. $\frac{7}{4}$

Answer 1

Answer: (C)

Let $AD$ be internal angle bisector of $\angle BAC$. Drop perpendicular from $B$ on $AD$ and 10439-40-41/st. line produce it to meet AC at B'.
Clearly, $\triangle AMB \cong \triangle AMB '$, so $B ^{\prime}$ is image of $B$ in line $AD$.
As slope of $AD$ is -3 , so slope of line $BB ^{\prime}=\frac{1}{3}=\tan \theta$ (say)

$\therefore$ Equation of $BB ^{\prime}$ in parametric form is $\frac{ x +1}{\frac{3}{\sqrt{10}}}=\frac{ y -3}{\frac{1}{\sqrt{10}}}= r$
$\therefore$ Co-ordinates of ' ' $^{\text {are }}\left(-1+\frac{3}{\sqrt{10}} \times 2 \times \frac{5}{\sqrt{10}}, 3+\frac{1}{\sqrt{10}} \times 2 \times \frac{5}{\sqrt{10}}\right)=(2,4)$
(As distance of line $AD$ from $B$ is $\frac{5}{\sqrt{10}}$ )
Clearly, equation of $AC$ is $y =2 x$ ....(1)
Also, equation of $AD$ is $3 x + y =5$ (given) .....(2)
$\therefore$ On solving (1) and (2), we get $A (1,2)$
As area $(\triangle ABC )=\frac{1}{2}(\sqrt{5})(2 \sqrt{5}) \sin 90^{\circ}=5$
so, area of triangle formed by joining midpoints of sides $B C, C A$ and $A B$ of $\triangle A B C$ $=\frac{1}{4} \operatorname{arca}(\triangle ABC )=\frac{5}{4}$.