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Question

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

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Solution

Step1: Simplification of given data
Let ABC be an equilateral triangle
In equilateral triangle all sides are equal
So, AB=BC=AC=2a
Given that the base of equilateral triangle i.e., BC lies along the y-axis
such that mid point of BC is O(0,0)
So, OB=OC=a
Coordinates of B=(0,a) and Coordinates of C=(0,a)
Also, Since O is the mid-point of BC
OA is the median of ABC
In equilateral triangle,
median & altitude coincide.
So, OA is the altitude of ABC
If OA is the altitude of ABC, then is OA perpendicular to BC

So, point A lies on xaxis as xaxis is perpendicular to yaxis.
Since point A lie on the xaxis
its y coordinate will be 0
Let the x coordinate be x
So, coordinates of point A=(x,0)

Step 2: Vertices of triangle
Now, AB=2a
Distance between (x,0) and (0,a) is 2a
(x2x1)2+(y2y1)2=2a
(x0)2+(0a)2=2a
(x)2+(a)2=2a
x2+a2=2a
Squaring both sides
((x2+a2))2=(2a)2
x2+a2=4a2
x2=4a2a2
x2=3a2
x=±3a2
x=±a3
So, point A=(x,0)
A=(a3,0) or (a3,0)

Therefore, vertices of the triangle are (0,a),(0,a) and (a3,0) or (0,a),(0,a) and (a3,0)

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