Midpoints of a triangle

Question

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

Answer 1

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

http://en.wikipedia.org/wiki/Heron%27s_Formula

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

Answer 2

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

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

Answer 3

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

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

Answer 4

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.

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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.