Question

a(-1,0) b(3,1), c(2,2) and d(-2,1) are the four points in a plane. show that ac bisect bd each other​

Answer 1

Two diagonals of a quadrilateral only bisect each other when the quadrilateral is a parallelogram. So, we simply have to prove that ABCD is a parallelogram.

We shall use the Distance Formula.

AB =

          $\sqrt{(3-(-1))^{2}+(1-0)^2 } \\\sqrt{17}$

BC =

        $\sqrt{(3-2)^2+(2-1)^2} \\\\\sqrt{2}$

CD =

        $\sqrt{(2-(-2))^2+(2-1)^2}\\\sqrt{17}$

AD =

        $\sqrt{((-1)-(-2))^2+(1-0)^2} \\\sqrt{2}$

As the opposite sides are equal, ABCD is a parallelogram.

Alternative proof

If AC and BD bisect each other, then their midpoints should coincide.

We will use the midpoint formula.

Midpoint of AC: $(\frac{2-1}{2}, \frac{2+0}{2})$

                           $(\frac{1}{2},1)$

Midpoint of BD:

                          $(\frac{3-2}{2},\frac{1+1}{2})\\(\frac{1}{2},1)$

The midpoint of AC and BD coincide. That means that AC and BD intersect. And their points of intersection bisect each other. In other words, AC and BD bisect each other.