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The exercise was:

Draw the triangle with vertices $A = (2;2)$, $B = (-1;3)$, and $C = (0;0)$. By regarding it as half of a parallelogram, explain why its area equals $$ \mathrm{area}(ABC) = \dfrac{1}{2} \det\begin{pmatrix} 2 & 2 \\ -1 & 3 \end{pmatrix}. $$

So I did explain this, explaining the area of the parallelogram as the determinant the give.

The 2nd question was:

Move the third vertex to $C = (1;-4)$ and justify the formula $$ \mathrm{area}(ABC) = \dfrac{1}{2} \det\begin{pmatrix} 2 & 2 & 1 \\ -1 & 3 & 1 \\ 1 & -4 & 1 \end{pmatrix}. $$

Now I'm having trouble explaining this. The solution says that the triangle has the same area, this provoked just a translation... But I don't understand why and how because now we have a different determinant... Also, how did they get the third column $(1,1,1)$? I'm confused... Thanks

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The usual way to calculate the area is the cross-product in $\mathbb R_3$.

Since we have $x_3=0$, we get $[a_1,a_2,0]\times[b_1,b_2,0]=[0,0,a_1b_2-a_2b_1]$. The length of this vector is $|a_1b_2-a_2b_1|=|det(\pmatrix{a_1&a_2\\b_1&b_2})|$.

Now, if $c\ne 0$, we have $[a_1-c_1,a_2-c_2,0]\times [b_1-c_1,b_2-c_2,0]=[0,0,(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)]$. And we have

$$|det(\pmatrix{a_1&a_2&1\\b_1&b_2&1\\c_1&c_2&1})|=|det(\pmatrix{a_1-c_1&a_2-c_2&0\\b_1-c_1&b_2-c_2&0\\c_1&c_2&1}|$$ , which coincides with the length of the vector $[0,0,(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)]$

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when three coordinates are given either we can arrange them row wise or column wise which results into an empty column or a row. so we just plug in $1$ in three places. in 1st question that $1$ dint matter as two entries were $0$. also we can have conditions where $det(A)=det(B)$ . if we have $3\times 3$ matrix $R_1[a b c],R_2[d e f],R_3[g h i]$. [ ] indicates rows. then its expansion is $a(ei-fh)-b(di-fg)+c(dh-eg)$ plug in values of three vertices from 1st question and then from 2nd you will get the same value. Hope its clear

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