Chapter 6.3, Problem 40HP

To determine

To calculate:

The locations of remaining vertices of parallelogram.

Explanation of Solution

Given information:

Intersecting point of diagonal: $\left(0,1\right)$

Vertex 1: $\left(2,4\right)$

Vertex 2: $\left(3,1\right)$

Calculation:

Graph each point as:

  

Consider,

The coordinates of the reaming vertices: $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ .

Here, diagonals of a parallelogram meet at the point $\left(0,1\right)$ . So, the midpoint formula is sued as:

  $\begin{array}{l}\left(\frac{{\mathrm{x}}_1+2}{2},\frac{{\mathrm{y}}_1+4}{2}\right)=\left(0,1\right)\\ \left(\frac{{\mathrm{x}}_2+3}{2},\frac{{\mathrm{y}}_2+1}{2}\right)=\left(0,1\right)\end{array}$

So,

  1. $\begin{array}{l}\frac{{\mathrm{x}}_1+2}{2}=0\\ {\mathrm{x}}_1+2=0\\ {\mathrm{x}}_1=-2\end{array}$ and $\begin{array}{l}\frac{{\mathrm{y}}_1+4}{2}=0\\ {\mathrm{y}}_1+4=0\\ {\mathrm{y}}_1=-4\end{array}$

  2. $\begin{array}{l}\frac{{\mathrm{x}}_2+3}{2}=0\\ {\mathrm{x}}_2+3=0\\ {\mathrm{x}}_2=-3\end{array}$ and $\begin{array}{l}\frac{{\mathrm{y}}_2+1}{2}=0\\ {\mathrm{y}}_2+1=0\\ {\mathrm{y}}_2=-1\end{array}$

Therefore, the coordinates of the remaining vertices are $\left(-3,1\right)$ and $\left(-2,-2\right)$ .