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Consider the triangle $(0,0),(4,0),(4,3)$. This has an area of 6 by the .5bh rule. Consider the following transformation

\begin{equation} \begin{bmatrix} -2 & -1 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 0 & 4 & 4\\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -8 & -11\\ 0 & 12 & 27 \end{bmatrix} \end{equation} Thus, we have a new triangle $(0,0) (-8,12) (-11,27)$.

Taking the determinant of the linear transformation yields $-7$ and, if we take the absolute value of this and multiply by $6$, the area of the preimage, we get $42$.

Yet .5bh yields a different result. If you draw the triangle, you get a skew triangle. You can divide this into two triangles and

$.5(3)(15) + .5(3)(12) = 40.5$

My Question:

Why am I getting inconsistent results from these two methods?

Stan Shunpike
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3 Answers3

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The area of the preimage triangle is $6$ (calculated from $\frac 12bh$ formula). The area of the image triangle is $42$ (calculated from $\frac 12bh$ formula). The determinant of your transformation matrix is $-7$. So your matrix transformed an area of $6$ to an area of $|6\cdot -7| = 42$.

So what's the problem here?

NOTE: If you're having problems calculating the area of the triangle from the points, just use this calculator.

Edit: Here's how I found it using the area formula:

The length of the side from $(0,0)$ to $(-11,27)$ is clearly just $\sqrt{11^2+27^2}$. That is the base. The height is then the distance from the line containing those two points to the point $(-8,12)$. Using the distance formula we can see that that is $$d(27x+11y=0,(-8,12))= \frac{|(27)(-8)+(11)(12)|}{\sqrt{27^2+11^2}}$$

Then $$\require{cancel}\frac 12bh = \frac 12\cancel{\sqrt{11^2+27^2}}\frac{|(27)(-8)+(11)(12)|}{\cancel{\sqrt{27^2+11^2}}} = \color{red}{42}$$

user275324
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  • I didn't get 42 from the $\frac 12bh$. As I explained, I got 40.5. If you can show what numbers you used for this, I might be able to figure out my error – Stan Shunpike Sep 30 '15 at 00:52
  • @StanShunpike Sorry. I had to make some dinner. I edited the calculation I used in. – user275324 Sep 30 '15 at 01:14
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If you cut the triangle vertically at the line $x=-8,$ you need to find the length of the vertical "base" which goes from the point $(-8,12)$ to the point on the line joining $(0,0)$ to $(-11,27),$ which has equation $y=-27x/11.$ This means the change in $y$ coordinates, which is the length of the base when cut this way, is $(-27/11)*(-8)-12=84/11.$ Then the two triangle areas are $(1/2)(84/11)*3=126/11$ for the left triangle and $(1/2)(84/11)*8=336/11,$ for a combined area of $42.$

The triangle could also be naturally cut using a horizontal slice, but again the length of the "base" is fractional, I think it was $34/9.4

coffeemath
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Coffeemath's answer explains your error. Here's an easy way to find the area of the skew triangle: the magnitude of the cross product of two vectors is the area of the parallellogram they span, so the area of your transformed triangle is $\frac12||\left<-8,12,0\right>\times \left<-11,27,0\right>||=\frac12{\tiny\left|Det\pmatrix{-8 & -11 \cr 12 & 27}\right|}=42=6\cdot 7$, as expected.

amd
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