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Question

The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

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Solution

Let ABC be the given equilateral triangle with side 2a

AB=BC=CA=2a

Since base, BC lies on the y-axis with mid-point at origin O.

OB=OC=a

So, the coordinates of point B are (0,a) and the coordinates of C are (0,a)

We know that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on x-axis.

So, let the coordinates of point A be (x,0)

On applying Pythagoras theorem ABC, we get

(AC)2=(OA)2+(OC)2

(2a)2=x2+a2

4a2a2=x2

x2=3a2

x=±3a

coordinates of point A are (±3a,0) or (3a,0)

Thus, the vertices of the given equilateral triangle are (0,a)(0,a) and (3a,0) or (0,a)(0,a) and (3a,0)

397108_418297_ans.png

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