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=}(y2-y1)/(x2-x1) \\ =(1-(-1))/(3-(-2)) \\ =(2)/(5) \\ \text{Slope of M'N'=}(y4-y3)/(x4-x3) \\ =(4-2)/(5-0) \\ =(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'=}(y3-y1)/(x3-x1) \\ =(4-(-1))/(5-(-2)) \\ =(3)/(2) \\ \text{Slope of NN'=}(y4-y2)/(x4-x2) \\ =(4-1)/(5-3) \\ =(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.