Question

The area of the triangle formed by the lines $x^2+4xy+y^2=0,x-y=4$ is

Answer 1

Since, it is given that $x-y=4----\left(1\right)$

represents one side of the triangle and $x^2+4xy+y^2$ represents the other two sides of the triangle.

$⟹x^2+2\times x\times 2\times y+(2y{)}^2-(\sqrt{3}y{)}^2)=0$

$⟹{(x+2y)}^2-{\left(\sqrt{3}y\right)}^2=0$

$⟹(x+2y-\sqrt{3}y)(x+2y+\sqrt{3}y)=0$

$⟹(x+2y-\sqrt{3}y)=0----\left(2\right)$

and

$(x+2y+\sqrt{3}y)=0----\left(3\right)$

which are the equation of the other two sides of the triangle.

Slope of equation (1)

$m_1=1$

slope of equation (2)

$m_2=-12+\sqrt{3}=-2-\sqrt{3}(2+\sqrt{3})(2-\sqrt{3})=-2-\sqrt{3}(4-3)=-(2+\sqrt{3})$

slope of the straight line (3)

$m_3=-12-\sqrt{3}=-2+\sqrt{3}(2+\sqrt{3})(2-\sqrt{3})=-2+\sqrt{3}(4-3)=-(2-\sqrt{3})$

Angle between the lines (1) and (2)

$⟹\theta ={tan}^{-1}\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)$

Putting the values of $m_1$ and $m_2,$ we get

$⟹\theta =tan-1\left(\sqrt{3}\right)=\frac{\Pi }{3}$

Similarly we can find the angle between the equations (1) and (3), which will be equal to theangle\quad between the equations (1) and (2) that is $\theta =\frac{\Pi }{3}$. Therefore, the given triangle is an equilateral triangle.Now the distance between two sides is given by $\frac{|c|}{\sqrt{a^2+b^2}}$ and height=$\frac{\sqrt{3}}{2}A$(where A is the side of the triangle)

$⟹\frac{|c|}{\sqrt{a^2+b^2}}=\frac{\sqrt{3}}{2}A$

$⟹\frac{2|c|}{\sqrt{3}\sqrt{a^2+b^2}}=A$

Area of the equilateral triangle $=\frac{\sqrt{3}}{2}\left(\frac{4c^2}{3(a^2+b^2)}\right)$

$⟹\frac{c^2}{\sqrt{3}(a^2+b^2)}$

$⟹\frac{(-1{)}^2}{\sqrt{3}(1+1)}$

$⟹\frac{1}{2\sqrt{3}}$ square units (Answer)