In order to prove that the diagonals of a square are congruent, first we need to find their length by using the Pythagorean Theorem:
[tex]h^2=a^2+b^2[/tex]Where h is the hypotenuse of the triangle (in this case the diagonal), a and b are the sides of the triangle. In the coordinate plane, we can see that each side measures 6 units. By replacing these values we can find the length of LN and KM as follows:
[tex]\begin{gathered} \bar{LN}=6^2+6^2 \\ \bar{LN}=36+36 \\ \bar{LN}=72 \\ \bar{LN}=\sqrt{72} \\ \bar{LN}=\sqrt[]{36\cdot2} \\ \bar{LN}=\sqrt[]{36}\cdot\sqrt[]{2} \\ \bar{LN}=6\sqrt[]{2} \end{gathered}[/tex]Now, for KM:
[tex]\begin{gathered} \bar{KM}=6^2+6^2 \\ \bar{KM}=36+36 \\ \bar{KM}=72 \\ \bar{KM}=\sqrt[]{36\cdot2} \\ \bar{KM}=\sqrt[]{36}\cdot\sqrt[]{2} \\ \bar{KM}=6\sqrt[]{2} \end{gathered}[/tex]Then, as the diagonals have the same measure they are congruent.
To find the slope, we can use the following formula:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]Were (x1,y1) and (x2,y2) are the coordinates of two points on the line.
The coordinates of K, L, M and N are:
K(2,1) and M(8,-5)
L(2,-5) and N(8,1).
The slope of KM is:
[tex]m=\frac{-5-1}{8-2}=\frac{-6}{6}=-1[/tex]And the slope of LN is:
[tex]m=\frac{1-(-5)}{8-2}=\frac{1+5}{6}=\frac{6}{6}=1[/tex]When two lines are perpendicular, the product of their slopes is equal to -1, then if m1
