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Area With Determinants - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
We’ve learned that we can use determinants to find the area of a parallelogram. But it turns out you can also use them to find the area of a triangle. Let’s take a look at a problem; let B equal the point (4, 3), A equal (7, -1), R equal (0, 2) and T equal (-3, 6). Now the quadrilateral BART happens to be a parallelogram. That’s going to be very helpful to us. Find the area of triangle BAT.
Now I’ve got B, A, R and T all plotted here. Here’s my parallelogram BART and here’s my triangle BAT. The thing we have to observe here, is that the area of triangle BAT is half the area of the parallelogram. So we know how to find the area of the parallelogram, we’ll just have to multiply that by ½. Let’s get started with that.
Let me write that down first. The area of triangle BAT is going to be ½ the area of parallelogram BART. Now if I wanted to find the area of this parallelogram, I’m going to need 2 vectors. Let’s say I focus on point B and the two vectors coming out of point B; vector BA and vector BT. So I’m going to need the components of those two vectors.
Now BA is the vector going from point B to point A, so it’s 7 minus 4, 3; -1 minus 3, -4. And BT is the vector going from B to point T, so this will be -3 minus 4, -7, and 6 minus 3, 3. So I have components for BA and BT.
Now remember the area formula for the parallelogram BART is going to be the absolute value, so ½, the absolute value, of the determinant followed by these two guys. So 3, -4, -7, 3. That’s ½ absolute value of, I have 9 minus 28. 9 minus 28 is -17. The absolute value of -17 is 17. I get ½ of 17 which is 8.5. So the area of triangle BAT is 8.5.
Remember we calculated here 17, that’s the area of the entire parallelogram. Whenever you draw a diagonal in a parallelogram, it divides the parallelogram into two triangles of equal area. So each of these little triangles is going to have an area of 8½.
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