Question

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

Answer 1

The length of each side of equilateral triangle is 2a and the base of an equilateral triangle lies along y- axis with its mid-point at the origin.

Suppose PQR be the given equilateral triangle of base side 2a . Since PQR is an equilateral triangle, the length of each side must be equal.

PQ=QR=RP=2a

Let QR be the base along the y axis, and O be the origin with vertices ( 0,0 ) such that,

QO=OR=a

It is clear that the coordinate of point Q is ( 0,−a ) and the point R is ( 0,a ) .

Now, for an equilateral triangle,the line joining themid-pointof any side with its opposite vertex is always perpendicular to that side.

It is given that the line passing through the mid-point is y- axis. So, the vertex P joining with other two vertices Q and R lies on y axis.

Apply Pythagoras theorem in triangle POQ .

PQ 2 = PO 2 + OQ 2

Substitute the values of PQ and OQ in the above formula.

( 2a ) 2 = PO 2 + a 2 PO 2 = ( 2a ) 2 − a 2 PO 2 =4 a 2 − a 2

Further simplify for the value of PO.

PO 2 =3 a 2 PO=± 3 a

The coordinate of point P is ( ± 3 a,0 ) .

The coordinate of point P is either ( 3 a,0 ) or ( − 3 a,0 ) .

The two possible equilateral triangles PQR are shown in the figure below.

Thus, the obtained vertices of triangle with side 2a lies along the y- axis with mid-point of the base lies at the origin, ( 3 a,0 ) , ( 0,−a ) , ( 0,a ) or ( − 3 a,0 ) , ( 0,−a ) , ( 0,a ) .