Q50 of 66 Page 7

The three vertices of a parallelogram ABCD are A (3, -4), B (-1, -3) and C (-6, 2). Find the coordinates of vertex D and find the area of ABCD.(CBSE 2013)

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Let the coordinates of the fourth vertex D be (x,y).


We know that-


Diagonals of a ∥gm bisect each other.


∴ Mid-point of diagonal AC ≡ Mid-point of diagonal BD



⇒ -3 = -1+x and -2 = -3+y


⇒ x = -2 and y = 1


Thus, the coordinates of the fourth vertex D is (-2,1).


∵ Area of a Δ ABC whose vertices are A(x1,y1), B(x2,y2) and C(x3,y3) is given by (1/2)|x1(y2-y3)+ x2(y3-y1)+ x3(y1-y2)| units2


∴ Area of given Δ ABC


= (1/2)|3(-3-2)+(-1)(2-(-4))+(-6)(-4-(-3))|


= (1/2)|3(-5)+(-1)(6)+(-6)(-1)|


= (1/2)|-15-6+6|


= (1/2)|-15|


= 15/2


Area can't be -ve so we will consider only magnitude.


∴ Area of Δ ABC = 15/2 sq. units


We know that-


Diagonals of a ||gm divides it into two triangles of equal area.


Thus, Area of ||gm ABCD = 2× 15/2 = 15 sq. units

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