The points $(4,2), (-1,-3)$, and $(-10,6)$ are the midpoints of the sides of triangle $ABC$. What is the area of triangle $ABC$?


closed as off-topic by user122283, Davide Giraudo, Sujaan Kunalan, user127096, TZakrevskiy Mar 30 '14 at 18:46



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    $\begingroup$ I've found the points of the triangle: (13,-7), (-5,11), (-15,1). $\endgroup$ – Ari Mar 30 '14 at 17:37
  • $\begingroup$ Yes - what else? $\endgroup$ – user122283 Mar 30 '14 at 17:37
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    $\begingroup$ Compute the area of the triangle whose vertices are the given three points and multiply by four. (Draw a picture of $\Delta ABC$ and connect the midpoints of the sides, partitioning this triangle into four congruent pieces.) $\endgroup$ – LeoTheKub Mar 30 '14 at 17:39
  • $\begingroup$ Should I use the distance formula to find the sides? $\endgroup$ – Ari Mar 30 '14 at 17:39
  • $\begingroup$ @Ari: you can use Determinant formula for area. demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant $\endgroup$ – mesel Mar 30 '14 at 18:08

Hint: The slope between $(4,2)$ and $(-1,-3)$ is 1. The slope between $(-1, -3)$ and $(-10, 6)$ is -1. Hence these two lines are perpendicular, and so we have a right triangle.

Hint: Find the area of the triangle.

Hint: Show that the median triangle is similar to the original triangle. What is the ratio?

Hint: Show that the area of the median triangle is $\frac{1}{4}$ the area of the original triangle.

Another approach: You have found the 3 vertices of the triangle. Show that 2 of those lines are perpendicular, and hence you have a right triangle. This makes it easy to find the area.

The reason why this works is motivated by the above.

  • $\begingroup$ What can the OP do if (s)he ever faces triangles that aren't right-angled?Just thought that this should be mentioned. $\endgroup$ – rah4927 Mar 30 '14 at 17:53
  • $\begingroup$ @rah4927 That would depend on the context in which OP received this question. Most of the time, Heron's isn't taught in school, which make make resorting to Heron's useless. I would prefer doing the shoelace formula, or even the "large rectangle minus several triangles" approach. $\endgroup$ – Calvin Lin Mar 30 '14 at 17:56
  • $\begingroup$ I tried the "large rectangle minus 3 triangles" method and got 180, which I believe is correct. Thank you! $\endgroup$ – Ari Mar 30 '14 at 18:01

Though computation intensive, you might want to check out Heron's formula now that you have the points.


Your task now is to find the side lengths from the given points!

  • $\begingroup$ Heron's is overkill in this situation. $\endgroup$ – Calvin Lin Mar 30 '14 at 17:46
  • $\begingroup$ No doubt! However, it never hurts to know a wide range of methods for calculating the area of a triangle. $\endgroup$ – Kaj Hansen Mar 30 '14 at 18:06

Hint: 1)Find the area of the medial triangle.

           2)What is the ratio of areas of the original triangle to the medial triangle?

Area of the whoole triangle is $4A$ where $A$ is the area of the triangle of midpoints.(triangle are similiar with ratio $2$) and

$$A={1\over 2}|det( \begin{bmatrix} 4 & 2 & 1 \\-1 & -3 & 1\\-10 & 6 & 1 \end{bmatrix}\quad|)=45$$

Thus, area of the big triangle is $4x45=180$

Instead of Heron Formula, determinant formula is more useful in analitic geometry.

If $(a,b),(c,d),(e,f)$ are the points then

$$A={1\over 2}|det( \begin{bmatrix} a & b & 1 \\c & d & 1\\e & f & 1 \end{bmatrix}\quad|)$$

