Question

Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments MN and M'N' have the same length.

Answer 1

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

$\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}$

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

$\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}$

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.