Question

The triangle ABC has medians AD, BE, CF. AD lies along the line $y=x+3$ , BE lies along the line $y=2x+4$, AB has length $60$ and angle $C=90$, then the area of ABC is

• A. $400$
• B. $200$
• C. $100$
• D. $50$

Answer 1

The correct option is A $400$

Given line

$y=x+3-----\left(1\right)$

$y=2x+4-----\left(2\right)$

By shiftiing centroid to origin

Thus now equation of median are $y=x$ and $y=2x$

Now the coordinates of A and B can be taken as $(a,a)$ and $B(b,2b)$

Using centroif formula $C(-a-b,-a-2b)$

Length of hypotenuse $AB=60$

$A{B}^2=(a-b{)}^2+(a-2b{)}^2$

$3600=a^2+b^2-2ab+a^2+4b^2-4ab$

$2a^2+5b^2-6ab=3600---\left(3\right)$

Slope of line AC from point $A,C$ is $m_{AC}=\frac{-a-2b-a}{-a-b-a}$

$m_{AC}=\frac{2a+2b}{2a+b}$

Slope of line BC from point $B,C$ is $m_{BC}=\frac{-a-2b-2b}{-a-b-b}$

$m_{BC}=\frac{a+3b}{a+2b}$

Angle $C=90$ means line AC and BC are perpendicular

Hence $m_{AC}{m}_{BC}=-1$

$\frac{2a+2b}{2a+b}\times \frac{a+3b}{a+2b}=-1$

$2a^2+6ab+2ab+6b^2=-2a^2-4ab-ab-2b^2$

$4a^2+8b^2+13ab=0--------\left(4\right)$

Solving both $\left(3\right)$ and $\left(4\right)$ equations we get

$ab=-\frac{800}{3}$

$Areaoftraingle=\frac{3}{2}\left(ab\right)=400$