# Thread: Area of a triangle

1. ## Area of a triangle

Use calculus to find the area of the triangle with the given vertices: (0,0) (10,1) (-1,2)

This completely blindsided me. What's going on here?

2. First, find the equations of the three lines that go through these points.
L1 -> the line going through (0, 0) and (-1, 2)
L2 -> the line going through (0, 0) and (10, 1)
L3 -> the line going through (-1, 2) and (10, 1)
Put all equations in slope-intercept form (y = mx + b). I'm going to call these three functions $L_1(x), L_2(x), L_3(x)$ respectively.

Draw the triangle. From x = -1 to x = 0, you can find the area of the part of the triangle to the left of the y-axis by finding the integral of the difference of the two functions L3(x) - L1(x):
$A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx \ldots$

Then, from x = 0 to x = 10, you find the area of the part of the triangle to the right of the y-axis by finding the integral of the difference of the two functions L3(x) - L2(x):
$\ldots + \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx$

So the complete integral will be
$A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx + \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx$

3. Do you have to use calculus?

You could just use the area of a triangle $\text{Area} =\frac{1}{2}ab\sin(C)$

4. I would break the problem up into two pieces: the area in the triangle between x = -1 and x = 0, and the area in the triangle between x = 0 and x = 10. For each of those triangles, you're going to have to compute the following:

$\int_{x_{\text{min}}}^{x_{\text{max}}}f_{\text{upp er}}(x)\,dx-\int_{x_{\text{min}}}^{x_{\text{max}}}f_{\text{low er}}(x)\,dx.$

Then you add the two together to find your total area. Make sense? I should say that this isn't the only way to solve this problem using calculus, but it's the way I would do it.

5. So the area is 12?

And yes, it has to be with calculus.

6. Can you elaborate on that?

7. I'm still not sure how to get it with this information...I've gotten closer but I don't know what the variables are in your work.

8. Originally Posted by bobsanchez
I'm still not sure how to get it with this information...I've gotten closer but I don't know what the variables are in your work.

You need to find the line segments mentioned in post #2. Are you firmilar with finding $y=mx+c$ given 2 points?

9. No...I mean, that reminds me of y = mx + b, but I don't know that specifically.

10. pickslides is talking about the same thing.

You'll need to find the slope between the two points for each line (this is m), and then you'll need to find the y-intercept (b). Although in two instances you are already given the y-intercept.

11. I didn't recognize it when it was added together (what form was that?) but I understand y = mx + c. I was just trying to clarify that that was indeed the process to go with.

12. Originally Posted by bobsanchez
I didn't recognize it when it was added together (what form was that?) but I understand y = mx + c.
Hi Bob, that was just a typo, all fixed now. You need to find the line for each of these from post #2

Originally Posted by eumyang
First, find the equations of the three lines that go through these points.
L1 -> the line going through (0, 0) and (-1, 2)
L2 -> the line going through (0, 0) and (10, 1)
L3 -> the line going through (-1, 2) and (10, 1)
Put all equations in slope-intercept form (y = mx + b). I'm going to call these three functions $L_1(x), L_2(x), L_3(x)$ respectively.

Draw the triangle. From x = -1 to x = 0, you can find the area of the part of the triangle to the left of the y-axis by finding the integral of the difference of the two functions L3(x) - L1(x):
$A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx \ldots$

Then, from x = 0 to x = 10, you find the area of the part of the triangle to the right of the y-axis by finding the integral of the difference of the two functions L3(x) - L2(x):
$\ldots + \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx$

So the complete integral will be
$A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx + \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx$
After that, integrate as explained in post #2.