Q19 of 50 Page 220

The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also find the centriod of the triangle.

Given; The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4).

Let he vertices be A(x1, y1, z1), B(x2, y2, z2) and A(x3, y3, z3) respectively.



x1 =2-x2, y1 =10-y2, z1 =-2- z2



x3 =-x2 , y3 =8-y2 , z3 =-4-z2




x2 = −1,


y2 = 6,


z2 = −7.


x1 = 2 − x2 = 3,


y1 = 10 − y2 = 4,


z1 = −4 – z2 = 5.


x3 = – x2 = 1,


y3 = 8 – y2 = 2,


z3 = −4 − z2 = 3.


A(−1, 6, −7), B(3, 4, 5), C(1, 2, 3) are the required vertices.


Centroid of a triangle is given by the average of the coordinates of its vertices or midpoint of sides.


Centroid is


More from this chapter

All 50 →