Chapter 5.1, Problem 18WE

To determine

To find: the third vertex of rectangle ABCD with the help of given information.

Answer to Problem 18WE

The third vertex of rectangle is $C(5,4)$

Explanation of Solution

Given information:

System information: $A(3,2),B(8,2),D(0,5)$

Calculation:

Quadrilateral rectangle ABCD is a parallelogram.

Diagonal of parallelogram bisect each other.

Consider that:

  $A(3,2),B(8,2),D(0,5)$

The midpoint formula here is,

Since, $AC=DB$

So,

  $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{x_3+x_4}{2},\frac{y_3+y_4}{2}\right)$

Put the value in the above equation.

  $\begin{array}{l}{x}_1=3;x_2=?\hfill \\ y_1=2;y_2=?\hfill \\ x_3=8;x_4=0\hfill \\ y_3=2;y_4=5\hfill \end{array}$

Substitute the values;

  $\left(\frac{3+x_2}{2},\frac{2+y_2}{2}\right)=\left(\frac{8+0}{2},\frac{2+5}{2}\right)$

Simplify the equations:

  $\frac{3+x_2}{2}=\frac{8}{2}$ and

  $\frac{3+y_2}{2}=\frac{7}{2}$

Simplify the equation,

  $\frac{3+x_1}{2}=\frac{8}{2}$

  $\begin{array}{l}{x}_1=8-3\hfill \\ x_1=5\hfill \end{array}$

And,

  $y_1=7-3$

  $y_1=4$

Hence, the third vertex of rectangle will be $C(5,4)$