Question

Find the internal angle of the triangle formed by the pair of straight lines $x^2-4xy+y^2=0$ and the straight line $x+y+4\surd 6=0$. Given the coordination of the vertices of the triangle so formed and also the area of the triangle.

Answer 1

$x^2-4xy+y^2=0$
or $y^2-4xy+4x^2=3x^2$
or $(y-2x{)}^2=(x\surd 3{)}^2$
$\therefore y-2x=\pm x\surd 3$
or $y=(2+\surd 3)x$, and $y=(2-\surd 3)x$, are the two
lines, the third line is $x+y+4\surd 6=0$.
Now $\tan \theta =\frac{2\surd (h^2-ab)}{a+b}$,
$a=b=1,h=-2$
$\therefore \tan \theta =2\frac{\surd (4-1)}{1+1}=\surd 3$.
$\therefore \theta ={60}^o$ and the corresponding vertex is clearly $(0,0)$.
Again if $\beta$ be the angle between the lines
$x+y+\surd 6=0$ and $y=(2+\surd 3)x$
Whose slopes are $-1$ and $2+\surd 3$ respectively, then

$\tan \beta =\frac{(-1)-(2+\surd 3)}{1+(2+\surd 3)(-1)}=\frac{-3-\surd 3}{-1-\surd 3}=\surd 3$

$\therefore \beta ={60}^o$

Since two angle are ${60}^o$ each therefore the third
angle is also ${60}^o$. Hence the triangle is
equilateral. Area of $△$

$p^2\tan {30}^o=\frac{1}{\surd 3}{p}^2=\frac{1}{\surd 3}.{\left(\frac{4\surd 6}{\surd (1+1)}\right)}^2$

$=16\surd 3sq$. units.
as $p$ is perpendicular from origin to the line
$x+y+4\surd 6=0$.