Topscore QR #14

[Questions]

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How do I find the answer?
Question is: IN a two dimensional coordinate system, point A=(3,-6), point B=(1,1), and point C=(7,6). Find the area of triangle of ABC

 

[Streetwolf]

How do I find the answer?
Question is: IN a two dimensional coordinate system, point A=(3,-6), point B=(1,1), and point C=(7,6). Find the area of triangle of ABC

Couple ways.

You could always find the line perpendicular to the base, find the height, and do 1/2 * b * h. This way will be long and annoying here.

You could use the law of cosines for the area. But here you'll get pretty bad values to work with.

Easiest way by far:

Using the determinant... it's easier if one of the coordinates is (0,0) so you're working with 2 vectors. To do that it's easiest to change point B to (0,0) and bring the other points down 1 and left 1. Then A = (2,-7) and C = (6,5). The determinant is found using the matrix with the top row (2,-7) and the bottom row (6,5). You take the top-left * bottom-right - top-right * bottom-left which is 2*5 - 6*-7 = 10 + 42 = 52. This is the area of the parallelogram formed from these 2 vectors. To find the area of the triangle, divide by 2 and get 26.

 

[Questions]

How did you know to use a matrix? Is that a topic covered in the DATs?
I don't yet quite understand your reasoning for the solution

 

[Streetwolf]

How did you know to use a matrix? Is that a topic covered in the DATs?
I don't yet quite understand your reasoning for the solution

Not sure if it's covered. Probably not. They probably wouldn't give this on the DAT. The attached picture explains my method.

The first picture is the triangle you have. The second picture is the triangle once I moved all 3 coordinates down one and left one.

The two arrows are the non-origin vertices of the triangle. I put the diagonal line connecting those two lines in so you could more easily visualize the triangle. Ignore that line and the dashed lines. The two lines that are left represent the vectors <6,5> and <2,-7>.

When you calculate the absolute value of the determinant of these two vectors, you get the area of the parallelogram they form. This is shown with the dashed lines (NOT the middle diagonal). Divide by 2 and you get the area of the triangle you are looking for.

You could do this with 3 non-origin points but it is more complicated.

Thus you could easily find the area of any triangle simply by placing one coordinate at the origin, adjusting the other 2 coordinates accordingly, and calculating the absolute value of the determinant of those 2 coordinates.

 

[Questions]

Thank you so much!!

 

[Streetwolf]

Okay the way you would be expected to do it is this way:

Pick any side. Let's go with the line between (3, -6) and (7, 6). The equation for this line has the slope of 3. So the equation is (y-6) = 3(x-7).

If this is the base, the height has to be perpendicular to it. So the slope of the height is -1/3. The height has to also touch the remaining point: (1,1).

The equation of this line is (y-1) = (-1/3)(x-1).

Now we need to know the distance of this line to get the distance of the height. So where does it touch the base? You need to find the intersection of those two lines:

Base: y = 3x - 15 (from the y-6 equation simplified)
Height: y = (-1/3)x + 4/3 (from the y-1 equation simplified)

Just set them equal: 3(x-5) = -1/3(x - 4)
-9(x-5) = x-4
-9x + 45 = x - 4
-10x = -49
x = 49/10
y = -3/10

You can see how much this already sucks.

So the endpoints of the height are (1,1) and (49/10, -3/10). The distance is:

sqrt of: (49/10 - 1)^2 + (-3/10 - 1)^2 = sqrt(1690/100)

The endpoints of the base are given to us as (3,-6) and (7,6) so the distance is sqrt(160).

The base is sqrt(160) or 4sqrt(10).
The height is sqrt(169/10) or 13sqrt(1/10)

Area = 1/2 b * h
Area = 1/2 (4sqrt(10) * 13sqrt(1/10))
Area = 1/2 * 4 * 13 * sqrt(10) * sqrt(1/10)
Area = 26 * sqrt(10/10)
Area = 26