Chapter 3.2, Problem 60PPS

To determine

To find: the point at which the diagonals of the quadrilateral intersects.

Answer to Problem 60PPS

  $x=6;y=\frac{7}{2}$ .

Explanation of Solution

Given:

The given graph is:

  

Calculation:

As The vertices of the quadrilateral have the coordinates. Then:

  $\begin{array}{l}A\left(2,2\right)\\ B\left(8,2\right)\\ C\left(10,5\right)\\ D\left(0,8\right)\end{array}$

Now to find the equation of the line passing through A, C:

  $\begin{array}{l}\frac{y-2}{5-2}=\frac{x-2}{10-2}\\ \frac{y-2}{3}=\frac{x-2}{8}\\ \Rightarrow 8\left(y-2\right)=3\left(x-2\right)\\ 8y-16=3x-6\\ \therefore 3x-8y=-10\end{array}$

Again, to find the equation of the line passing through B, D:

  $\begin{array}{l}\frac{y-2}{8-2}=\frac{x-2}{0-8}\\ \frac{y-2}{6}=\frac{x-2}{-8}\\ \Rightarrow -8\left(y-2\right)=6\left(x-2\right)\\ -8y+16=6x-12\\ \therefore 3x+4y=32\end{array}$

The intersection of the diagonals is the solution of the system:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}3x-8y=-10\\ 3x+4y=32\\ ---\end{array}}\\ -12y=-42\\ \therefore y=\frac{-42}{-12}=\frac{7}{2}\end{array}$

Substitutes the value of y in the above equation. Then:

  $\begin{array}{l}3x+4y=32\\ 3x+4\left(\frac{7}{2}\right)=32\\ 3x+14=32\\ 3x=18\\ \therefore x=6\end{array}$

Hence, the point at which the diagonals of the quadrilateral intersects is $x=6;y=\frac{7}{2}$ .