Question

If a $∆ABC$ has vertices$(0,0),(11,60)$ and $(91,0)$. If the line $y=kx$ cuts the triangle into two triangles of equal area, then $k$is equal to

• A. $\frac{30}{51}$
• B. $\frac{4}{7}$
• C. $\frac{7}{4}$
• D. $\frac{30}{91}$

Answer 1

The correct option is A

$\frac{30}{51}$

Explanation for correct option:

Find the value of $k$:

Given, $∆ABC$ has vertices $(0,0),(11,60),(91,0)$

And line $y=kx$ cuts the triangle into two triangles of equal area.

Median pass through $C$ will bisect the triangle into two equal area triangle.

Mid point of $AandB$ will lie on $y=kx.$

Mid point $\left(\frac{11+91}{2},\frac{60+0}{2}\right)=(51,30)$

Therefore

$$\begin{gathered} 30=k51 \\ \Rightarrow k=\frac{30}{51} \end{gathered}$$

Hence, the correct option is (A).