Question

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

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

Answer 1

Answer: (C)

$\frac{10}{3}$

Answer 2

The area of triangle formed by the lines $ax^{2} + 2hxy + by^{2} = 0$ and $lx + my + n = 0$is given by $\left| \frac{n^{2}\sqrt{h^{2} - ab}}{am^{2} - 2hlm + bl^{2}} \right|$

Here $a = 4,b = - 9,h = - \frac{9}{2},l = 1,m = 0,n = - 2$,

then area of triangle

=$\left| \frac{( - 2)^{2}\sqrt{\left( \frac{- 9}{2} \right)^{2} - 4 \times \frac{- 9}{2}}}{- 9 \times (1)^{2}} \right|$=$\left| \frac{4\sqrt{\frac{81}{4} + \frac{36}{2}}}{- 9} \right| = \left| \frac{- 30}{9} \right| = \frac{10}{3}$