Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).
Let the co-ordinates of P and R be (a,b) and (c,d) and coordinates of Q are (3, 2)
By midpoint formula.
x =
, y = ![]()
(2 , - 1) is the mid-point of PQ.
∴ 2 =
and -1 = ![]()
∴ a = 1 and b = -4
∴ Coordinates of P are (1, -4)
(1 , 2) is the mid-point of QR.
∴ 1 =
and 2 = ![]()
∴ c = -1 and d = 2
∴ Coordinates of P are (-1, 2)
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
=
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area of ∆PQR =
| 3( − 4 – 2 ) + 2( − 1 – 1) + 1( 2 − 4) |
=
| − 18 − 4 − 2 |
= 12 sq. units
Hence the area of ∆PQR is 12 sq. units