Before we start the lesson, we will go over when to use analytic geometry and when not to. Analytic geometry rarely produces pretty solutions and is prone to arithmetic errors and the like in the algebra. Thus, we encourage you to not jump to analytic geometry to solve ever geometry problem. Now, this isn't to say that you should never use analytic geometry. It is a very useful tool to have and can solve many problems easily that may have otherwise been very hard. Analytical geometry is especially effective when there are lots of lengths given, several right angles, and the intersection of lines. However, even on most problems that do seem to fit these conditions, we encourage to try to find a non-analytic (synthetic) solution first, as these solutions are often much quicker.
We will start by introducing a theorem which we will not prove. There is a proof using induction and the cross product which is too advanced for the curriculum. However, this theorem is quite necessary for many problems that use "coordinate bashing" as their solution.
We will start by introducing a theorem which we will not prove. There is a proof using induction and the cross product which is too advanced for the curriculum. However, this theorem is quite necessary for many problems that use "coordinate bashing" as their solution.
The Shoelace Formula
This formula provides a way to find the area of any polygon given its vertices in the order they are connected. The best way to explain it is to show a visual.
![Picture](/uploads/1/1/2/8/112821707/cmathg11_1_orig.png)
If we let the sum of the products along the blue diagonals is b and the sum of the products along the red diagonals is r, then the area of the polygon with the that is made by connecting each point to the one directly below it is equal to \(\frac{1}{2}\mid{b-r} \mid\). Also, note that we rewrote the first coordinate of the polygon at the bottom.
To make this clearer, we will go over an example. Say that we had a triangle with vertices (1,-2), (7,8), and (1,1). Note that if this weren't a triangle we would have to draw a quick sketch to get the order of the points, but for a triangle any ordering will be correct because each point is connected to the other two. Thus ,we write out the points in order.
(1,-2)
(7,8)
(1,1)
(1,-2)
From here we get the area to be \(\frac{1}{2}\mid (1\cdot 8 +7\cdot 1 +1\cdot -2)-(-2\cdot 7 +8\cdot 1 + 1\cdot 1)\mid=9\).
To make this clearer, we will go over an example. Say that we had a triangle with vertices (1,-2), (7,8), and (1,1). Note that if this weren't a triangle we would have to draw a quick sketch to get the order of the points, but for a triangle any ordering will be correct because each point is connected to the other two. Thus ,we write out the points in order.
(1,-2)
(7,8)
(1,1)
(1,-2)
From here we get the area to be \(\frac{1}{2}\mid (1\cdot 8 +7\cdot 1 +1\cdot -2)-(-2\cdot 7 +8\cdot 1 + 1\cdot 1)\mid=9\).
Sample Problem 1. Find the area of triangle ABF given that E is the midpoint of CB.
In order to show how to coordinate bash a problem, the best way is to show an example. This problem's solution will contain many important techniques that you should remember.
First of all, we will start by setting our coordinate axis. We let A=(0,0), B=(1,0), and D=(1,0). Thus, E=(1,0.5). Next, using the shoelace formula, we know we can find the desired area if only we had the coordinates of F. However, we know that we can do this by finding the intersection of two lines. We can note that line AE has the equation \(y=\frac{x}{2}\). We can also note that line BD has the equation \(y=-x+1\). Thus, setting the two lines equal and solving we get that their intersection is \((\frac{2}{3},\frac{1}{3})\). Now, we could use the shoelace formula to find the triangle's area, but in general we should look if there is a quicker way first. In this case, now that we have the y coordinate of F, we know that AB has length 1 and the height draw to AB has length 1/3. Thus, triangle ABF has area \(\frac{1}{2}(1\cdot\frac{1}{3})=\boxed{\frac{1}{6}}\). An important note is that while coordinate bashing, we can use other formulas or some synthetic (non-coordinate) techniques to reduce our calculations.
As you saw in this problem, the key to analytical geometry is to find the coordinates of points. you saw several useful methods of how to do this in the problem, but we will create a more extensive list below for you to use when solving problems.
How to find the coordinates of a point
Finally, when you are coordinate bashing you can choose any point to be the origin and any lines as the x and y axis. Thus, you should make sure to choose an origin an axis that make calculations easier (more specifically, choose a point that would make other points easier to find using the methods to find points provided above).
First of all, we will start by setting our coordinate axis. We let A=(0,0), B=(1,0), and D=(1,0). Thus, E=(1,0.5). Next, using the shoelace formula, we know we can find the desired area if only we had the coordinates of F. However, we know that we can do this by finding the intersection of two lines. We can note that line AE has the equation \(y=\frac{x}{2}\). We can also note that line BD has the equation \(y=-x+1\). Thus, setting the two lines equal and solving we get that their intersection is \((\frac{2}{3},\frac{1}{3})\). Now, we could use the shoelace formula to find the triangle's area, but in general we should look if there is a quicker way first. In this case, now that we have the y coordinate of F, we know that AB has length 1 and the height draw to AB has length 1/3. Thus, triangle ABF has area \(\frac{1}{2}(1\cdot\frac{1}{3})=\boxed{\frac{1}{6}}\). An important note is that while coordinate bashing, we can use other formulas or some synthetic (non-coordinate) techniques to reduce our calculations.
As you saw in this problem, the key to analytical geometry is to find the coordinates of points. you saw several useful methods of how to do this in the problem, but we will create a more extensive list below for you to use when solving problems.
How to find the coordinates of a point
- Find the equations of two lines (or any other shape) that the point is on and compute the intersection of the lines to find the point.
- Find a point that the point you are given is directly above or directly horizontal to for which you know the distance between these two points. Further, if you know that a point is some number to the left or right and some number up or down from another point whose coordinates you know, you can easily find that points coordinates.
- If a point is the midpoint of a line segment, then its coordinates are the average of those of that segment's endpoints.
Finally, when you are coordinate bashing you can choose any point to be the origin and any lines as the x and y axis. Thus, you should make sure to choose an origin an axis that make calculations easier (more specifically, choose a point that would make other points easier to find using the methods to find points provided above).
Exercises
- Find the centroid of a triangle with coordinates A=\((1,2)\),B= \((3,4)\), and C=\((3,6)\).
- Find the Area of triangle ABE given that E is the intersection of the quarter-circle centered at B through A and line BD.
Solution 1
We know that the centroid is the intersection of the medians. Thus, we simply need to find the formulas for the medians. The midpoint of AB is (2,3), so we get that the median to C has the equation y=3x-3. The midpoint of BC is (3,5), so the equation for the median to A is y=3x/2+1/2. These two medians intersect at \(\boxed{(7/3,4)}\).
Try to generalize this for a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), and \((x_3,y_3)\) to see if you can find a general formula.
Try to generalize this for a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), and \((x_3,y_3)\) to see if you can find a general formula.
Solution 2
To find E, we will just find the intersect between line BD and circle B though A keeping in mind that E has an x value between 0 and 1. note that the circle has equation \((x-1)^2+y^2=1\). Line BD has equation y=-x+1. Thus, they intersect at \(x=1\pm \frac{1}{\sqrt{2}}\). However, because we know x is less than 1, we see that E is at \(1\pm \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\). Thus, the area of triangle AEB is \(\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\cdot 1=\boxed{\frac{\sqrt{2}}{4}}\).