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.


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 $3y + 5x = 0$


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 $3y + 5x = 0$

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.