 Hello and welcome to the session, let us discuss the following question. It says the base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle. Let us now move on to the solution. We are given an equilateral triangle say abc and the base lies along the y-axis. So we are given an equilateral triangle say abc since it is an equilateral triangle all the sides are equal that is ab is equal to ac is equal to vc and the length is given to be 2a and we are given that the base lies along the y-axis and origin is the midpoint of the base. Since origin is the midpoint of bc therefore ob is equal to oc is equal to a since bc has length 2a and o is the midpoint therefore ob is equal to oc is equal to half of 2a that is a. Now since b lies along y-axis therefore x coordinate must be 0 and ob has length a therefore v has coordinates as 0a and similarly c has coordinates as 0 minus a. Now we have to find the coordinates of a. Now in triangle aob ab square is equal to ob square plus oa square since it is a right triangle and this implies oa square is equal to ab square minus ob square. Now ab is 2a therefore ab square is 2a square and ob is a therefore ob square is a square and this is equal to 4a square minus a square which is equal to 3a square. Now taking the square root on both sides we have oa is equal to plus minus root 3a. Now since a lies along x-axis therefore coordinate of y must be 0 and the coordinate of a will be root 3a 0 if a lies to the right side of y-axis and coordinates of a will be minus root 3a if a lies to the left side of y-axis. Hence the vertices of triangles are 0a 0 minus a and minus root 3a 0 or 0a 0 minus a root 3a 0. We can see that vertices of b and c remains same in both the cases. So this completes the question hope you enjoyed this session goodbye and take care.