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 $3x+y=5$.
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, $△AMB\cong △AM{B}^{\prime }$, so $B^{\prime }$ is image of $B$ in line $AD$.
As slope of $AD$ is -3 , so slope of line $B{B}^{\prime }=\frac{1}{3}=\tan \theta$ (say)

$\therefore$ Equation of $B{B}^{\prime }$ 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 $y=2x$ ....(1)
Also, equation of $AD$ is $3x+y=5$ (given) .....(2)
$\therefore$ On solving (1) and (2), we get $A(1,2)$
As area $(△ABC)=\frac{1}{2}(\sqrt{5})(2\sqrt{5})\sin 90^{\circ }=5$
so, area of triangle formed by joining midpoints of sides $BC,CA$ and $AB$ of $△ABC$ $=\frac{1}{4}\mathrm{arca}(△ABC)=\frac{5}{4}$.