# 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.