Question

The area of the triangle formed by the line ${4\mathrm{x}}^2-9x\mathrm{y}-{9\mathrm{y}}^2=0$ and $\mathrm{x}=2$ is

• A. $2$
• B. $3$
• C. $\frac{10}{3}$
• D. $\frac{20}{3}$

Answer 1

The correct option is C

$\frac{10}{3}$

Explanation for the correct option:

Step 1: Plotting the graph

Given,

${4\mathrm{x}}^2-9x\mathrm{y}-{9\mathrm{y}}^2=0\left(\mathrm{i}\right)$

$\mathrm{x}=2$ or $\mathrm{x}-2=0$

Solving $\left(\mathrm{i}\right)$ by bi-factorization method,

$\begin{array}{rcl}4{\mathrm{x}}^2-9\mathrm{xy}-3\mathrm{xy}+3\mathrm{xy}-9{\mathrm{y}}^2& =& 0\\ 4\mathrm{x}(\mathrm{x}-3\mathrm{y})+3\mathrm{y}(\mathrm{x}-3\mathrm{y})& =& 0\\ (4\mathrm{x}+3\mathrm{y})(\mathrm{x}-3\mathrm{y})& =& 0\end{array}$

Thus, now we have three linear equations,

$\begin{array}{rcl}4\mathrm{x}+3\mathrm{y}& =& 0\\ \mathrm{x}-3\mathrm{y}& =& 0\\ \mathrm{x}-2& =& 0\end{array}$

Visualizing the three equations on a graph.

We can see from that diagram that the vertices of the required triangle will be:

$(0,0),(2,\frac{-8}{3}),(2,\frac{2}{3})$

Step 2: Calculating the area of the triangle

Therefore, the area of the triangle is given by,

$\text{Area}=\frac{1}{2}\left|\begin{array}{l}{x}_1{y}_{1}1\\ x_2{y}_{2}1\\ x_3{y}_{3}1\end{array}\right|$

$$\begin{gathered} =\frac{1}{2}\left|\begin{array}{ccc}2& -\frac{8}{3}& 1\\ 2& \frac{2}{3}& 1\\ 0& 0& 1\end{array}\right| \\ =\frac{1}{2}\left(1\left(2\times \frac{2}{3}-2\times \frac{-8}{3}\right)\right) \\ =\frac{1}{2}\left(\frac{4}{3}+\frac{16}{3}\right) \\ =\frac{1}{2}\times \frac{20}{3} \\ =\frac{10}{3} \end{gathered}$$

Hence, option (C) is the correct answer.