### Putting geometry to work

Analytic geometry or coordinate geometry is geometry with numbers. In analytic geometry, vertices and special points have coordinates — (x, y) in the 2D plane, (x, y, z) in 3D space, and so on. Curves are represented by equations. For example, the graph of x2 + y2 = 1 is a circle in the x-y plane.

We manipulate these coordinates and equations to change geometric figures, explore their properties and create new forms. Many of the foundational ideas in this section have been covered in other sections on distance, slope, midpoint and conic sections.

Photo: General Electric Aviation

#### Ideas → Reality

Analytic geometry is how we translate engineering, architectural, artistic and other ideas into a language that builders, machinists and machines can use to physically create the things of which we dream.

The Boeing 777 aircraft, ubiquitous in the skies now as a passenger carrier, was the first such plane that was entirely built and tested on computers before a prototype was manufactured.

Imagine trying to manufacture the complexly-shaped turbofan blades of the GE turbine engine (left). Each contains complicated curves in each of the three dimensions that must be precisely formed in hard metals that are likely difficult to work with.

Parts like these are made by programmed machines, and that programming is mathematical coordinates and equations that specify every feature of the part, translated from an engineer's mind to his/her computer to the machine.

In this section, we will learn some analytic geometry by example. Below are several representative problems that I hope will help you get a good feel for the kinds of things that can be done with analytic geometry. These problems usually involve many steps, so I'll try to lay them out that way.

### Example 1

If you're able to solve this kind of problem, you know a lot about linear equations, slope, distance, midpoint and basic coordinate geometry:

Find the equation of a line perpendicular to, and passing through the midpoint of, the segment that has endpoints (-2, -3) and (4, 4).

Solution: Here are the steps we'll take to solve this problem.

1. It's helpful to plot the two given points on a graph (a sketch is all that's needed), and draw the segment.

2. Then we'll need the midpoint of that segment, for which we'll employ the midpoint formula (average the x-coordinates and average the y-coordinates)

3. Next we'll need to calculate the slope of our segment, then the slope of a line perpendicular to it is the negative reciprocal of that.

4. finally, we have the slope of our new line and a point through which it passes, so we can find its equation and put it into a pleasing form.

#### 2. Find the midpoint

The midpoint of a segment can be found just by finding the average of the x-coordinates and the average of the y-coordinates.

I strongly suggest that you keep fractions as fractions in these calculations. Often, doing so will help you to recognize when simple numerators or denominators cancel to simplify your work.

#### 3. Calculate the slope

The slope of the segment with endpoints (-2,-3) and (4, 4) is calculated here. As usual, the slope of a line is just the change in y (Δ y) divided by the change in x (Δ x) between two points.

The slope of our segment is 7/6, so the slope of a line perpendicular to it has the negative reciprocal slope, m = -6/7.

#### 4. Find the equation of the new line

Now that we have a point (the midpoint of the original segment) and a slope (what we just calculated above), we have enough information to find the equation of our perpendicular line.

We can rearrange the slope-intercept form of the equation of a line, y = mx + b, to

$$b = y - mx$$

and plug in our slope and midpoint (x, y) to find the y-intercept, b:

$$b = \frac{1}{2} = \frac{6}{7}(1) = -\frac{5}{14}$$

Put it all together to get the equation of our line:

$$y = -\frac{6}{7}x - \frac{5}{14}$$

or $\:\:12x + 14y = -5$

In that last step, I just multiplied everything through by 14 to get rid of the fractions. It makes for a neater result. Here's a picture of the two lines.

### Example 2

Now let's apply a similar kind of problem solving to a different problem, finding the centroid of a triangle. The centroid is the center of area. Think of it like this: If the triangle was made of something like metal with a continuous thickness, the centroid would be its center of balance — the triangle would balance perfectly at this point.

Find the centroid of the triangle with vertices at (0, 4), (4, 2) and (-3, -2).

Solution: The centroid of a triangle is located at the intersection of the three medians. We only need to find two, but it's a good idea to find all three because the arithmetic in these problems can get tricky and that third median will confirm whether our centroid location is correct.

Recall that the centroid of a triangle is a line segment drawn from one vertex to the midpoint of the opposite side. Our strategy will be:

1. Find the midpoints of all three sides

2. Using the midpoint and corresponding vertex, find the equation of each of the three medians
3. Solve the equations of the medians simultaneously to find the location of the centroid, the one point that solves the equations of all of the medians.

#### Finding the midpoints

The midpoints of each line segment are found using the midpoint formula, as we did in the last problem.

When the coordinates of the triangle are as simple as these, it's easy to find midpoints by inspection — midpoints are just the ordered pair: (average of the x-coordinates, average of the y-coordinates). The resulting midpoints are shown in the graph.

#### Finding the equations

The process for finding equations of each of the three medians are shown in the table below. It is most convenient to express them in standard form in preparation for solving for the values of x and y that satisfy all three equations — the centroid.

#### The centroid

The first two equations are easy to solve together. They can be added to eliminate y and arrive at x = 1/3. Then re substitution of x into either equation gives y = 4/3.

The location of the centroid (1/3, 4/3) can be substituted into the third equation and confirmed. Here is a picture of the centroid location.

### Example 3

Here's an analytic geometry problems that will hone your skills with conic sections and call on your best algebra skills.

Find the point(s) of intersection of the curves defined by these equations:

$$\frac{x^2}{4} + \frac{(y - 2)^2}{4} = 1 \; \; \color{#E90F89}{\text{and}} \; \; \frac{x^2}{9} + \frac{y^2}{4} = 1$$

Solution: The equation on the left is a circle of radius r = 2 and centered at (0, 2):

$$\frac{x^2}{4} + \frac{(y - 2)^2}{4} = 1$$

The other is an ellipse centered at the origin with a major axis (along x) of 3 units and a minor axis (along y) of two units:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

Here's a picture of the two:

We can use the circle to solve for $x^2:$

$$x^2 + \frac{9}{4} y^2 = 1 \; \rightarrow \; x^2 = 9 - \frac{9}{4} y^2$$

Now substitution of that value of x2 for x2 in the ellipse equation gives

\begin{align} 9 - \frac{9}{4} y^2 + (y - 2)^2 &= 4 \\[5pt] 9 - \frac{9}{4} y^2 + y^2 - 4y + 4 &= 4 \\[5pt] -\frac{5}{4} y^2 - 4y + 9 &= 0 \end{align}

Now complete the square to solve this equation (find the roots):

\begin{align} -\frac{5}{4} y^2 - 4y &= -9 \\[5pt] y^2 + \frac{16}{5} y &= \frac{36}{5} \\[5pt] y^2 + \frac{16}{5} y + \left( \frac{8}{5} \right)^2 &= \frac{180}{25} + \frac{64}{25} \\[5pt] \left( y + \frac{8}{5} \right)^2 &= \frac{244}{25} \\[5pt] y = \frac{-8 ± \sqrt{244}}{5} &= \frac{-8 ± 2 \sqrt{61}}{5} \end{align}

Now no intersection of these curves can exist below y = 1, so we can discard the negative (minus) solution, giving the y-coordinate of our solutions as

$$\frac{-8 + 2 \sqrt{61}}{5} = 1.524$$

Now our formula for x, from above, was

$$x^2 = 9 - \frac{9}{4} y^2,$$

giving

$$x = ± \sqrt{9 - \frac{9}{4} \cdot 1.524^2},$$

or     $x = ± 2.106$

These are our two intersection points: $(-2.106, 1.524)$ and $(2.106, 1.524).$

It's worth considering just how two closed figures, like ellipses, might intersect. If you think about it, you'll come up with these possibilities:

You can see that two ellipses can intersect not at all, once, twice, three or four times. Note that some of the intersections are points of tangency. These will appear as double roots in your calculations, while lack of intersection will result in no solution. There are many paths through calculations like this, but choose wisely to limit the complexity of the calculation. Complex calculations are rich with possibilities for small errors.

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