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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.

Main Page:Geometry:Straight Lines-II