Question

If A(1,2), B(2,1), C(3,-4) are the vertices of ؈MABCD, find the coordinates of D.

Answer 1

As shown in the figure diagonal AC and BD intersect at midpoint O in parallelogram.

So, midpoint of AC and BD will be same

midpoint of AC = $(\frac{1+3}{2} ,\frac{2-4}{2}) =(2,-1)$

hence, midpoint of BD = $(\frac{x_{0}+2}{2} ,\frac{y_{0}+1}{2} ) =(2,-1)$

$x_{0} =2 ,y_{0}= -3$

Answer 2

we know, diagonals of parallelogram intersect at midpoint.

in ABCD, AC and BD are diagonals .

so, midpoint of AC = midpoint of BD

Let D = (x , y)

midpoint of AC = {(1 + 3)/2 , (2 - 4)/2 } = (2, -1)

midpoint of BD = {(2 + x)/2, (1 + y)/2 }

so, 2 = (2 + x)/2

=> 2 × 2 = (2 + x)

=> x = 2

-1 = (1 + y)/2

=> -1 × 2 = (1 + y)

=> -2 - 1 = y

=> y = -3

hence, D = (2,-3)