Question

Area of the triangle formed by the lines $y^2-9xy+18x^2=0$ and y=9 is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The lines represented by $y^2-9xy+18x^2=0$ are 6x - y = 0 and 3x - y = 0 and third line is y = 9.

Therefore, coordinates of vertices of triangle are given by A(0,0); B(3,9) and c$(\frac{3}{2},9)$

Hence, area of $△ABC=\frac{1}{2}$ $\left|\begin{array}{ccc}0& 0& 1\\ 3& 9& 1\\ \frac{3}{2}& 9& 1\end{array}\right|=\frac{27}{4}sq.units.$

Aliter: Applying the formula given in the theory part, the required area is

$\frac{(-9{)}^2\sqrt{(\frac{9}{2}{)}^2}-18}{18\times 1+9\times 0\times 1+1\times 0}=\frac{81}{18}\sqrt{\frac{81}{4}-18}$

= $\frac{81}{18}\times \frac{3}{2}=\frac{27}{4}$sq.units.