Chapter 0.1, Problem 55E

To determine

To find: Draw the three vertices and give co-ordinate of the missing vertices.

Answer to Problem 55E

The vertices are $\left(-1,4\right),\left(5,2\right),(-1,-2)$ ..

Explanation of Solution

Given information:

The given vertices are $\left(-1,1\right),\left(2,0\right)$ and $\left(2,3\right)$ .

Calculation:

The given vertices are $\left(-1,1\right),\left(2,0\right)$ and $\left(2,3\right)$ .

Find the slope of the line passing through the point $\left(-1,1\right),\left(2,3\right)$ .

The formula of slope is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute the values $x_1=-1,y_1=1,x_2=2,y_2=3$ .

  $\begin{array}{l}m=\frac{3-1}{2-(-1)}\\ m=\frac{2}{3}\end{array}$

The equation of line passing through $(-1,1)$ and with slope $\frac{2}{3}$ is:

  $y-y_1=m\left(x-x_1\right)$

Substitute the found values in the above equation:

  $\begin{array}{l}y-1=\frac{2}{3}\left(x+1\right)\\ 3y-3=2(x+1)\\ 3y-3=2x+2\\ 3y=2x+5\\ y=\frac{2}{3}x+\frac{5}{3}\mathrm{.....}(1)\end{array}$

Since the figure is of parallelogram therefore the line parallel to $y=\frac{2}{3}x+\frac{5}{3}$ has the same slope as this line.

Now find the equation of line with slope $\frac{2}{3}$ and point $\left(2,0\right)$ .

Apply the formula $y-y_1=m\left(x-x_1\right)$ take $x_1=2,y_1=0$

  $\begin{array}{l}y-0=\frac{2}{3}\left(x-2\right)\\ 3y=2x-4\end{array}$

Now find the equation of line which passes through $\left(-1,1\right),\left(2,0\right)$ .

Find the slope of the line passing through the point $\left(-1,1\right),\left(2,0\right)$ .

The formula of slope is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute the values $x_1=-1,y_1=1,x_2=2,y_2=0$ .

  $\begin{array}{l}m=\frac{0-1}{2-(-1)}\\ m=\frac{-1}{3}\end{array}$

Since we need to find the parallelogram so the line parallel to the line which passes through $\left(-1,1\right),\left(2,0\right)$ will have same slope.

The equation of line passing through $(2,3)$ and with slope $\frac{-1}{3}$ will be found with the formula $y-y_1=m\left(x-x_1\right)$ .

Substitute $x_1=2,y_1=3,m=\frac{-1}{3}$

  $\begin{array}{l}y-3=\frac{-1}{3}\left(x-2\right)\\ 3y-9=-1\left(x-2\right)\\ 3y=-x+2+9\\ 3y=-x+11\mathrm{....}(2)\end{array}$

Now draw the lines (1) and $(2)$ and the given points on the graph it will gives three parallelogram show below:

  

The vertices of the other parallelograms are $\left(-1,4\right),\left(5,2\right)$ .