Question

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

• A. $\frac{27}{4}$ sq units
• B. $27$ sq units
• C. $\frac{27}{2}$ sq units
• D. None of these

Answer 1

The correct option is A

$\frac{27}{4}$ sq units

Explanation for the correct option:

Step 1. Draw the curves $y^2-9xy+18x^2=0$ and $y=9$:

Given pair of straight line $y^2-9xy+18x^2=0$

Let the above line represents lines $y=m_{1}x$ and $y=m_{2}x$

$\Rightarrow$ $(y-m_{1}x)(y-m_{2}x)=y^2-9xy+18x^2$

$\Rightarrow$$y_2-(m_1+m_2)x+m_1{m}_2=y^2-9xy+18x^2$

Step 2. On comparing both sides, we get

$m_1+m_2=9$ and $m_1{m}_2=18$

$m_1=3,6$and $m_2=6,3$

So, the equations are $y=3x,y=6x$

Now, point of intersection of $y=3x,y=6x,y=9$ are $(0,0),(3,9),\left(\frac{3}{2},9\right)$

$\therefore$Area of the triangle $=\frac{1}{2}\times \frac{3}{2}\times 9$

$=\frac{\mathbf{27}\mathbf{}}{\mathbf{4}}$ sq units

Hence, Option ‘A’ is Correct.