To find the exact coordinates of the point where the diagonals intersect find the equation of each diagonal and then find the solution (where the lines cross each other)
Slope:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]For the diagonal that passes for points D ( 0 , 0 ) and B ( 8 , 4)
Slope:
[tex]m=\frac{4-0}{8-0}=\frac{4}{8}=\frac{1}{2}[/tex]y-intrpcet (b):
[tex]\begin{gathered} y=mx+b \\ 4=\frac{1}{2}(8)+b \\ 4=4+b \\ 4-4=b \\ 0=b \end{gathered}[/tex]Equation:
[tex]y=\frac{1}{2}x[/tex]For the diagonal that passes for points A ( 1, 5) and C (6 , 2)
Slope:
[tex]m=\frac{2-5}{6-1}=-\frac{3}{5}[/tex]y-intercept (b):
[tex]\begin{gathered} y=mx+b \\ 5=-\frac{3}{5}(1)+b \\ 5=-\frac{3}{5}+b \\ 5+\frac{3}{5}=b \\ \frac{25+3}{5}=b \\ \\ \frac{28}{5}=b \end{gathered}[/tex]Equation:
[tex]y=-\frac{3}{5}x+\frac{28}{5}[/tex]-----------------------------
You have the next sytem of equations:
[tex]\begin{gathered} y=\frac{1}{2}x \\ \\ y=-\frac{3}{5}x+\frac{28}{5} \end{gathered}[/tex]To solve:
1. Substitute the value of y in the second equation for the one in the first equaion:
[tex]\frac{1}{2}x=-\frac{3}{5}x+\frac{28}{5}[/tex]2. Solve for x:
[tex]\begin{gathered} \frac{1}{2}x+\frac{3}{5}x=\frac{28}{5} \\ \\ \frac{5x+6x}{10}=\frac{28}{5} \\ \\ \frac{11}{10}x=\frac{28}{5} \\ \\ x=(\frac{28}{5})(\frac{10}{11}) \\ \\ x=\frac{280}{55} \\ \\ x=\frac{56}{11} \end{gathered}[/tex]3. Use the value of x to find the value of y:
[tex]\begin{gathered} y=\frac{1}{2}x \\ \\ y=\frac{1}{2}(\frac{56}{11}) \\ \\ y=\frac{56}{22} \\ \\ y=\frac{28}{11} \end{gathered}[/tex]Then, the point where the diagonas intersect is (56/11, 28/11) approximately (5.091,2.545)[tex]\begin{gathered} (\frac{56}{11},\frac{28}{11}) \\ \\ (5.091,2.545) \end{gathered}[/tex]