Riemann surface
matlab

Show that the points $(2,1),(4,0)$, and $(5,7)$ are vertices of a right triangle.

Solution
The three points are plotted in Figure 8. Using the Distance Formula, you can find the lengths of the three sides as follows.
\begin{aligned} & d_1=\sqrt{(5-2)^2+(7-1)^2}=\sqrt{9+36}=\sqrt{45} \ & d_2=\sqrt{(4-2)^2+(0-1)^2}=\sqrt{4+1}=\sqrt{5} \ & d_3=\sqrt{(5-4)^2+(7-0)^2}=\sqrt{1+49}=\sqrt{50} \end{aligned}
Because
$$\left(d_1\right)^2+\left(d_2\right)^2=45+5=50=\left(d_3\right)^2$$
you can conclude by the Pythagorean Theorem that the triangle must be a right triangle.

Find the midpoint of the line segment joining the points $(-5,-3)$ and $(9,3)$.

Solution
Let $\left(x_1, y_1\right)=(-5,-3)$ and $\left(x_2, y_2\right)=(9,3)$.
\begin{aligned} & \text { Midpoint }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \quad \text { Midpoint Formula } \ & =\left(\frac{-5+9}{2}, \frac{-3+3}{2}\right) \quad \text { Substitute for } x_1, y_1, x_2 \text {, and } y_2 \text {. } \ & =(2,0) \quad \text { Simplify. } \ & \end{aligned}
The midpoint of the line segment is $(2,0)$, as shown in Figure 9.

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