Chapter 5.2, Problem 25PSC

To determine

To Prove: Quadrilateral obtained by the mid-points of a side of RECT is a parallelogram..

Explanation of Solution

Given:

  

Concept Used:

Parallelogram is a quadrilateral of which opposite sides are parallel and equal in length.

  $\text{Coordinates of mid-point of side having end-points}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

  $\text{Slope of line joining the points}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\frac{y_2-y_1}{x_2-x_1}$

Proof:

As we know that

  $\text{Coordinates of mid-point of side having end-points}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Coordinates of mid-points of RECT are

  $\begin{array}{l}\text{Coordinates of mid-point of side }\overline{RE}=\left(\frac{0+0}{2},\frac{0+4}{2}\right)=\left(0,2\right)\\ \text{Coordinates of mid-point of side }\overline{EC}=\left(\frac{0+12}{2},\frac{4+4}{2}\right)=\left(6,4\right)\\ \text{Coordinates of mid-point of side }\overline{CT}=\left(\frac{12+12}{2},\frac{4+0}{2}\right)=\left(12,2\right)\\ \text{Coordinates of mid-point of side }\overline{RT}=\left(\frac{0+12}{2},\frac{0+0}{2}\right)=\left(6,0\right)\end{array}$

Thus, the figure obtained by joining the mid-points of RECT as

  

Now we will prove that ABCD is a parallelogram using the coordinates of A, B, C and D.

  $\begin{array}{l}\text{Since, slope of line joining the points}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\frac{y_2-y_1}{x_2-x_1}\\ \text{Slope of side }\overline{AB}=m_1=\frac{2-0}{0-6}=-\frac{1}{3}\\ \text{Slope of side }\overline{BC}=m_2=\frac{4-2}{6-0}=\frac{1}{3}\\ \text{Slope of side }\overline{CD}=m_3=\frac{2-4}{12-6}=-\frac{1}{3}\\ \text{Slope of side }\overline{DA}=m_4=\frac{2-0}{12-6}=\frac{1}{3}\\ \therefore m_1=m_3\\ \Rightarrow \text{Slope of side }\overline{AB}=\text{Slope of side }\overline{CD}\\ \text{and }{m}_2=m_4\\ \Rightarrow \text{Slope of side }\overline{BC}=\text{Slope of side }\overline{DA}\\ \text{Thus, ABCD is a parallelogram}\text{.}\end{array}$