Question

If two vertices of a parellelogram are $(3,2)$ and $(-1,0)$ and the diagonals intersect at $(2,-5)$, then the other two vertices are:

• A. $(1,-10),(5,-12)$
• B. $(1,-12),(5,-10)$
• C. $(2,-10),(5,-12)$
• D. $(1,-10),(2,-12)$

Answer 1

Answer: (B)

$(1,-12),(5,-10)$

Let the given vertices be $A(1,-12)B(5,-10)C(a,b);D(x,y)$
We know that the diagonals of a parallelogram bisect each other. So,the midpoint of AC $=(2,-5)$ and of BD $ = (2,-5) $.

Mid point of two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is calculated by the formula
$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Hence, midpoint of $AC=(2,-5)$
$=>(\frac{1+a}{2},\frac{-12+b}{2})=(2,-5)$
$=>\frac{1+a}{2}=2;\frac{-12+b}{2}=-5$
$=>1+a=4;-12+b=-10$
$a=3;b=2$
Hence, C $=(3,2)$
And, midpoint of $BD=(2,-5)$
$=>(\frac{5+x}{2},\frac{-10+y}{2})=(2,-5)$
$=>\frac{5+x}{2}=2;\frac{-10+y}{2}=-5$
$=>5+x=4;-10+y=-10$
$x=-1;y=0$
$=>D=(-1,0)$