Answer :

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


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