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

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.

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.

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):

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