Question

Show that the lines $x^2-4xy+y^2=0$ and $x+y=10$ contain the sides of an equilateral triangle. Find the area of the triangle.

Answer 1

We find the joint equation of the pair of line $OA$ and $OB$ through origin, each making an angle of ${60}^o$ with $x+y=10$ whose slope is $-1$

Let $OA\left(orOB\right)$ has slope $m$

$\therefore$ its equation is $y=-mx...\left(1\right)$

Also, $\tan {60}^o=\left|\frac{m-(-1)}{1+m(-1)}\right|$

$\therefore \sqrt{3}=\left|\frac{m+1}{1-m}\right|$

Squaring both sides, we get,

$3=\frac{(m+1{)}^2}{(m-1{)}^2}$

$\therefore 3(1-2m+m^2)=m^2+2m+1$

$\therefore 3-6m+3m^2=m^2+2m+1$

$\therefore 2m^2-8m+2=0$

$\therefore m^2-4m+1=0$

$\therefore \left(\frac{x}{y}\right)-4\left(\frac{x}{y}\right)+1=\mathrm{0...}\left[By\right(1\left)\right]$

$\therefore y^2-4xy+x^2=0$

$\therefore x^2-4xy+y^2=0$ is the joint equation of the two lines through the origin each making an angle of ${60}^o$ with $x+y=10$

$\therefore x^2-4xy+y^2=0$ and $x+y=10$ from a triangle $OAB$ which is equilateral

let seg $OM$ perpendicular line $AB$ whose question is $x+y=10$

$\therefore MO=\left|\frac{-10}{\sqrt{1+1}}\right|+=5\sqrt{2}$

$\therefore$ are of equilateral $△OAB=\frac{(OM{)}^2}{\sqrt{3}}=\frac{(5\sqrt{2}{)}^2}{\sqrt{3}}$

$=\frac{50}{\sqrt{3}}squnits.$