Question

The area of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is square units. is

• A. $\frac{1}{12}$
• B. $\frac{1}{2\sqrt{5}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{1}{12}$

$x^2-3xy+2y^2=0$

$\Rightarrow (x-y)(x-2y)=0$

Hence three sides are

$x-y=0⟶\left(i\right)$

$x-2y=0⟶\left(ii\right)$

$x+y+1=0⟶\left(iii\right)$

Solving $\left(i\right)$, $\left(ii\right)$ & $\left(iii\right)$

three vertices are $A(0,0),B(\frac{-2}{3},\frac{-1}{3}),C(\frac{-1}{2},\frac{-1}{2})$

$AB=\sqrt{\frac{1}{9}+\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$

$BC=\sqrt{\frac{1}{36}+\frac{1}{36}}=\frac{\sqrt{2}}{6}$

$AC=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$\therefore$ area using heron's formula

$\Delta =\sqrt{S(S-AB)(S-BC)(S-CA)}$

$=\frac{1}{12}{unit}^2$