Question

The area of the triangle formed by the lines $4x^2-9xy-9y^2=0$ and $x=2$ is equal to

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

Answer 1

Answer: (C)

$\frac{10}{3}$ sq units

$4x^2-9xy-9y^2=0$ and $x=2$

$4x^2-9xy-3xy+3xy-9y^2=0$

$4x(x-3y)+3y(x-3y)=0$

$(4x+3y)(x-3y)=0$

thus the three equations of the line are :-

$4x+3y=0$

$x-3y=0$

$x-2=0$

The equation for the bounded area(which is a triangle ) is:

$=\frac{\begin{array}{cccc}{x}_1& y_1& 1& \\ x_2& y_2& 1& \\ x_3& x_3& 1& \end{array}}{2}$

$=\frac{\begin{array}{cccc}0& 0& 1& \\ 2& \frac{2}{3}& 1& \\ 2& -\frac{8}{3}& 1& \end{array}}{2}$

Upon solving matrix we get :-

The bounded area $=1(2\times \frac{-8}{3}-2\times \frac{2}{3})$

$=\frac{-16}{4}-\frac{4}{3}$

$=\frac{-20}{3\times 2}$

$=\left|\frac{-10}{3}\right|$

$=\frac{10}{3}$sq unit