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Area of a Triangle

Petrus

Well-known member
Feb 21, 2013
739
In an orthogonal cordinate system determine the area of the triangle with vertices in \(\displaystyle (-4,1)\), \(\displaystyle (1,4)\) and \(\displaystyle (-5,10)\) There is prob many way to solve it so go ahead with your method:) I made a hint for vector method.



Hint
The area of a triangle is half of the area of parallelogram
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Re: Area

The vector cross product, I believe (or the magnitude thereof) gives the area of the corresponding parallelogram. Hence, divide by two to get the answer. I get

$$[ \langle -4,1,0 \rangle- \langle 1,4,0 \rangle] \times [ \langle -4,1,0 \rangle- \langle -5,10,0 \rangle]= \langle -5,-3,0 \rangle \times \langle 1,-9,0 \rangle$$
$$= \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}\\ -5 &-3 &0 \\ 1 &-9 &0\end{matrix} \right| = \hat{k}((-5)(-9)-(-3)(1))= \hat{k}(45+3)=48 \hat{k}.$$
Hence, the area of the triangle is $24$ units squared.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Area

Using the formula I developed in this topic, we find:

\(\displaystyle A=\frac{1}{2}|(-5+4)(4-1)-(1+4)(10-1)|=\frac{48}{2}=24\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: Area

In an orthogonal cordinate system determine the area of the triangle with vertices in \(\displaystyle (-4,1)\), \(\displaystyle (1,4)\) and \(\displaystyle (-5,10)\) There is prob many way to solve it so go ahead with your method:) I made a hint for vector method.



Hint
The area of a triangle is half of the area of parallelogram
If you get really desperate, you could work out the length of each side of the triangle by evaluating the distance between each set of points, and then use Heron's Formula.
 

TheEmptySet

New member
Mar 15, 2012
8
Re: Area

The good old fashion method. The slope between the points (-4,1) and (1,4) is \(\displaystyle \frac{3}{5}. \)So the slope of a line perpendicular to it is \(\displaystyle -\frac{5}{3}\)This gives the two equations \(\displaystyle y-1=\frac{3}{5}(x+4)\) the equation of the line perpendicular is \(\displaystyle y-10=-\frac{5}{3}(x+5)\) Solving the system for the point of intersection gives \(\displaystyle \left(-\frac{13}{17},\frac{50}{17} \right)\) The length of each line segment can now be calculated \(\displaystyle b=\int_{-4}^{1}\sqrt{1+(\frac{3}{5})^2}dx=\frac{24}{17} \sqrt{34}\\ h=\int_{-5}^{-\frac{13}{17}}\sqrt{1+(-\frac{5}{3})^2}dx=\sqrt{34}\\\) This gives the area of the triangle \(\displaystyle A=\frac{1}{2}bh=24\)