Question

Using the section formula, show that the points
A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are vertices of a parallelogram taken in order.

Answer 1

Diagonals of a paralleogram bisect each other.
So, the midpoint of the diagonal AC must coincide with the midpoint of diagonal BD.
Using section formula we can write,

$(l{x}_2+m{x}_1)/(l+m),(l{y}_2+m{y}_1)/(l+m)$

l = 1 and m = 1

The midpoint of the diagonal AC divides the line AC in the ratio 1:1

= $\frac{1\left(2\right)+1\left(1\right)}{(+1}$ , $\frac{1\left(7\right)+1\left(0\right)}{1+1}$

= $\frac{2+1}{2}$ , $\frac{7+0}{2}$

= $\frac{3}{2}$, $\frac{7}{2}$ ------- (1)

The midpoint of the diagonal BD divides the line BD in the ratio 1:1

= $\frac{1(-2)+1\left(5\right)}{1+1}$ , $\frac{1\left(4\right)+1\left(3\right)}{1+1}$

= $\frac{-2+5}{2}$ , $\frac{4+3}{2}$

= $\frac{3}{2}$, $\frac{7}{2}$ ------- (2)

So, the diagonals bisect each other at a common point.
Hence, the given vertices form a parallelogram.