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Find the area of the triangle in the plane $R^2$ bounded by the lines $y = x$, $y = -3x+8$, and $3y + 5x = 0$

I know that I can find the area of the triangle by taking the half of the area of the parallelogram the points make. But I don't know how to convert those equations to points so I can take the vectors and calculate it's determinant.

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Find the points $(x_i, y_i)$ at which each pair of equations intersect. There are three such pairs, so three points of intersection, which will be the vertices of the triangle. From the vertices $(x_i, y_i)$, you can determine the two vectors you need, which you can use as the columns of the matrix for which the absolute value of the determinant, multiplied by 1/2, will give you area.

Given $\;y = x$, $y = -3x+8$, and $3y + 5x = 0$

Example:

Vertex 1: At what point does $y = x$ and $y = -3x + 8$ intersect? When $x = -3x + 8$. Solving for $x$, gives us $x = 2$, which in this case, will also equal $y$.

So vertex 1 is $(2, 2)$. Proceed in a similar manner to determine:

Vertex 2: where $y = x$ and $3y + 5x = 0$.

Vertex 3: where $y - -3x + 8$ and $3y + 5x = 0$

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  • $\begingroup$ Thank you. A very clear explanation. I solve the problem, the area is 16 $\endgroup$ – Randolf Rincón Fadul Apr 10 '13 at 21:06
  • $\begingroup$ You're very welcome! $\endgroup$ – Namaste Apr 10 '13 at 21:08
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Or from the vertices, you can apply the formula:

$$Area = \dfrac12 \left|\begin{array}{cccc} x_1 && x_2 && x_3 && x_1 \\ y_1 && y_2 && y_3 && y_1 \end{array}\right|$$

If you know how to use it.

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