Question

Find the area of the triangle formed by the lines $4{x^2} - 9xy - 9{y^2} = 0$ and $x = 2$.
A. $2$
B. $3$
C. $\dfrac{{10}}{3}$
D. $\dfrac{{20}}{3}$

Answer 1

Hint: Find the two straight lines by factoring the given equation of a pair of straight lines. You will get two straight lines. Then solve the three equations of the three lines and obtain the vertices of the triangle. After that use the formula of finding the area of a triangle.

Formula Used:
The area of a triangle having the vertices $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ is given by the determinant $\dfrac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 &1 \\ x_3 & y_3 & 1\end{vmatrix}$

Complete step by step solution:
The given equation of the pair of straight lines is $4{x^2} - 9xy - 9{y^2} = 0.....(i)$.
Factorize the expression on the left hand side of the equation.
$4{x^2} - 9xy - 9{y^2}\\ = 4{x^2} - 12xy + 3xy - 9{y^2}\\ = 4x\left( {x - 3y} \right) + 3y\left( {x - 3y} \right)\\ = \left( {x - 3y} \right)\left( {4x + 3y} \right)$
Now, from equation $(i)$, we get
$\left( {x - 3y} \right)\left( {4x + 3y} \right) = 0$
$ \Rightarrow x - 3y = 0$ or $4x + 3y = 0$
So, equations of the two straight lines obtained from the equation of the pair of straight lines are
$x - 3y = 0.....(ii)$ and $4x + 3y = 0.....(iii)$
And another one given straight line is $x = 2.....(iv)$
Now, solve the equations $(ii),(iii),(iv)$to obtain the coordinates of the vertices of the triangle.
At first solve equations $(ii),(iii)$
Adding the equations $(ii),(iii)$, we get $5x = 0 \Rightarrow x = 0$ and hence $y = 0$
So, the point of intersection of the straight lines $(ii),(iii)$ is $O\left( {0,0} \right)$
Now, solve equations $(ii)$ and $(iv)$
Putting the value of $x$ from equation $(iv)$in equation $(ii)$, we get $2 - 3y = 0 \Rightarrow y = \dfrac{2}{3}$
So, the point of intersection of the straight lines $(ii)$ and $(iv)$ is $A\left( {2,\dfrac{2}{3}} \right)$
Now, solve equations $(iii)$ and $(iv)$
Putting the value of $x$ from equation $(iv)$ in equation $(iii)$, we get $8 + 3y = 0 \Rightarrow y = - \dfrac{8}{3}$
So, the point of intersection of the straight lines $(iii)$ and $(iv)$ is $B\left( {2, - \dfrac{8}{3}} \right)$
$\therefore $ The vertices of the triangle are $O\left( {0,0} \right)$, $A\left( {2,\dfrac{2}{3}} \right)$ and $B\left( {2, - \dfrac{8}{3}} \right)$.
Find the area of the triangle using the given formula.
The area is $\dfrac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 2 & {\dfrac{2}{3}} & 1 \\ 2 &{-\dfrac{8}{3}} & 1\end{vmatrix}$ square units
Expand the determinant with respect to the first row.
$ = \dfrac{1}{2}\begin{vmatrix} {2 \times \left( { - \dfrac{8}{3}} \right) - 2 \times \dfrac{2}{3}} \end{vmatrix}\\ = \dfrac{1}{2}\begin{vmatrix} { - \dfrac{{16}}{3} - \dfrac{4}{3}} \end{vmatrix}\\ = \dfrac{1}{2}\begin{vmatrix} { - \dfrac{{20}}{3}} \end{vmatrix}\\ = \dfrac{1}{2} \times \dfrac{{20}}{3}\\ = \dfrac{{10}}{3}$
$\therefore $The area is $\dfrac{{10}}{3}$ square units.

Option ‘C’ is correct

Note: Whenever an equation of a pair of straight lines is given, you need to factorize it to get the two equations of the two straight lines. You should be aware of the formula for finding the area of a triangle by using a determinant and taking the absolute value of the area because the area of anything must be positive.