Question

Consider a $△ABC$ in the $xy$-plane with vertices $A=(0,0),B=(1,1)$ and $C=(9,1)$. If the line $x=a$ divides the triangle into two parts of equal area, then $a$ equals:

• A. $3$
• B. $3.5$
• C. $4$
• D. $4.5$

Answer 1

The correct option is A $3$

Area of the triangle $=\frac{1}{2}\times (9-1)\times 1=4$ sq units.

Since we require half area, the area of the new triangle must be $2$ sq units.

The equation of line from $(0,0)$ to $(9,1)$ is $9y=x$

So, at $x=a$, the height of the triangle would be $(1-\frac{a}{9})$

Corresponding base would be $(9-a)$

Thus, area is given by $\frac{1}{2}\times (9-a)\times (1-\frac{a}{9})=2$

$\Rightarrow 9-a-a+\frac{a^2}{9}=4$

$\therefore a^2-18a+45=0$

$\therefore a=3,6$