Students will learn several properties of sigma notation, as seen here and here. Then students will practice applying the properties to prove identities involving sigma notation. Here's an interesting problem. Conclude by giving your students this challenge.

Concude by giving your students this challenge:

There is a point \((x_0, y_0)\) on a line with slope \(m.\) At a distance of \(d,\) along this line, there is another point \((x, y).\) Express \(x,\) and \(y\) in terms of \(x_0,\) \(y_0,\) \(m,\) and \(d.\)

Here's one solution: Think about our situation geometrically. There is a line, and some point on that line. The point is also the center of a circle with radius \(d.\) We are interested in finding the two points where the line intersects the circle. From the equation of a circle:

$$(x - x_0)^2 + (y - y_0)^2 = d^2$$

And the slope formula tells us that

$$m = \dfrac{y - y_0}{x - x_0}$$

From this, we obtain

$$m(x - x_0) + y_0 = y$$

Substituting into our first equation, we get

$$(x - x_0)^2 + m^2(x - x_0)^2 = d^2$$

Which can be factored as

$$(x - x_0)^2\left(1 + m^2\right) = d^2$$

Rearranging, we get

$$(x - x_0)^2 = \dfrac{d^2}{\left(1 + m^2\right)}$$

Taking the square root of both sides, then adding \(x_0\) to both sides, gives us

$$x = x_0 \pm \dfrac{d}{\sqrt{1 + m^2}}$$

Now let's follow the same procedure for \(y.\) Starting from the slope formula, once again, we get

$$\dfrac{1}{m}(y - y_0) + x_0 = x$$

Substituting that into our circle equation, just as before, then making the same deductions, just as before, we get

$$\begin{align}
& (y - y_0)^2 + \dfrac{1}{m^2}(y - y_0)^2 = d^2 \\[1em]
& (y - y_0)^2\left(1 + \dfrac{1}{m^2}\right) = d^2 \\[1em]
& (y - y_0)^2 = \dfrac{d^2}{1 + \dfrac{1}{m^2}} \\[1em]
& (y - y_0)^2 = \dfrac{m^2d^2}{m^2 + 1} \\[1em]
& y = y_0 \pm \dfrac{md}{\sqrt{m^2 + 1}}
\end{align}$$

In summary

$$\begin{align}
& x = x_0 \pm \dfrac{d}{\sqrt{1 + m^2}} \\[1em]
& y = y_0 \pm \dfrac{md}{\sqrt{m^2 + 1}}
\end{align}$$

TODO: Diagram of the geometric situation.

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