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  • If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle.

If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle.

Answers (1)

Let ABC be an equilateral triangle, BC is base, and altitude from A on BC meets at mid-point D.

Given:Equation of the base BC is x+y=2 

\sin 60^{0}=\frac{AD}{AB}

\frac{\sqrt{3}}{2}=\frac{AD}{AB}

AD=\frac{\sqrt{3}}{2}AB

  Distance of point (x1,y1) from the equation Ax+By+C=0 

  d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}

 Now, length of perpendicular from vertex A(2,-1) to the line x+y=2  

AD=\frac{\left | 1*2+1\left ( -1 \right ) -1 \right |}{\sqrt{\left ( 1 \right )^{2} +\left ( 1 \right )^{2}}}

\frac{\sqrt{3}}{2}AB=\left | \frac{2-1-2}{\sqrt{2}} \right |=\frac{1}{\sqrt{2}}

  Squaring both the sides, we get \frac{3}{4}AB^{2}=\frac{1}{2}

AB^{2}=\frac{4}{3}*\frac{1}{2}=\frac{2}{3}

AB=\sqrt{\frac{2}{3}}

Posted by

infoexpert22

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