Geo3.8

Show that the lines $x^2+16xy-11y^2=0\,$ form an equilateral triangle with the line $2x+y+1=0\,$ and find its area.

Given equation is $x^2+16xy-11y^2=0\,$

$(x+8y)^2-75y^2=0,(x+8y)^2-(5\sqrt{3})^2=0\,$

$[x+8y+5\sqrt{3}y][x+8y-5\sqrt{3}y]=0\,$

$[x+(8+5\sqrt{3})y][x+(8-5\sqrt{3})y]=0\,$

Let these two lines be $L_1\equiv 0,L_2\equiv 0\,$ respectively.

Let $L_3\equiv 2x+y+1=0\,$

Angle between L1,L2 is $\cos\alpha=\frac{a+b}{\sqrt{(a-b)^2+4h^2}}=\frac{1-11}{\sqrt{(1+11)^2+4(8)^2}}=\frac{-1}{2}\,$

Now $\alpha=60^\circ\,$ or $120^\circ\,$

Similarly,the angle between L1,L3 is $\cos\beta=\frac{a_1 a_2+b_1 b_2}{\sqrt{(a_1^{2}+b_1^{2})(a_2^{2}+b_2^{2})}}\,$

$\cos\beta=\frac{10+5\sqrt{3}}{2(10+5\sqrt{3}}=\frac{1}{2}\,$

$\beta=60^\circ\,$ or $120^\circ\,$

Therefore the three lines $L_1,L_2,L_3\,$ form a triangle.

Since the sum of the angles in any triangle is $180^\circ\,$

Since the angles that we found are $60^\circ,60^\circ\,$

Hence the three lines form an equilateral triangle.

Let p be the perpendicular distance from the vertex(0,0) to the line $2x+y+1=0\,$

Therefore, $p=\frac{|0+0+1|}{\sqrt{4+1}}=\frac{1}{\sqrt{5}}\,$

Area of the triangle is $\frac{p^2}{\sqrt{3}}=\frac{1}{5\sqrt{3}}\,$ units.