Question

Prove that the lines represented by $3x^2-8xy-3y^2=0$ and $x+2y=3$ form the sides of an isosceles right angled triangle.

Answer 1

$3x^2-8xy-3y^2=0$

$\Rightarrow 3x^2-9xy+xy-3y^2=0$

$\Rightarrow 3x(x-3y)+y(x-3y)=0$

$\Rightarrow (3x+y)(x-3y)=0$

Lines are $3x+y=0$ and $x-3y=0$

(Refer to Image)

Vertices are $(0,0)$

$x+2y=3$ $y=-3x$

$x-3y=0$ $x+2(-3x)=3$

$x=3y$ $-5x=3$

$3x+2y=3$ $x=-3/5,y=-3(-3/5)=9/5$

$5y=3$

$y=3/5$ $x=3\left(3/5\right)$

$x=9/5$

$|AB|=|AC|$

$△ABC$ is isosceles triangle.

$|AB{|}^2=\sqrt{\left(\frac{3}{5}\right)+{\left(\frac{9}{5}\right)}^2}=\frac{90}{25}$

$\Rightarrow |AC{|}^2=\frac{90}{25}$

$\Rightarrow A{B}^2+A{C}^2=\frac{180}{25}$

$\Rightarrow |BC|=\sqrt{{\left(\frac{12}{5}\right)}^2+{\left(\frac{6}{5}\right)}^2}=\sqrt{\frac{144+36}{25}}=\sqrt{\frac{180}{25}}$

$\Rightarrow B{C}^2=\frac{180}{25}$

$\Rightarrow A{B}^2+A{C}^2=B{C}^2$

$\therefore$ $△ABC$ is an isosceles right angles triangle.