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

Area of the triangle formed by the line $y^2-9 x y+18 x^2=0$ and $y=9$ is
  1. $\frac{27}{3}$ sq. units
  2. $\frac{27}{2}$ sq. units
  3. $\frac{27}{4}$ sq. units
  4. 27 sq. units

Solution

$\begin{aligned} & y^2-9 x y+18 x^2=0 \\ & \therefore(y-3 x)(y-6 x)=0 \end{aligned}$ Thus three lines forming triangle are $y=3 x, y=6 x, y=9$ Their point of intersections are $(0,0),(3,9),\left(\frac{3}{2}, 9\right)$ $\therefore$ Area of triangle $=\frac{1}{2} \times \frac{3}{2} \times 9$ $=\frac{27}{4}$ sq. units

Asked in: MHT CET 2021 (22 Sep Shift 1)

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