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Finding the area of a triangle formed by 3 points in the plane

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MarkFL

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Feb 24, 2012
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Suppose we have 3 points in the plane given by:

$\displaystyle (x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)$

and we wish to find the area of the triangle whose vertices are at these points.

We may let the base b of the triangle be the line segment between the first two points, and the altitude h of the triangle will be the perpendicular distance from the third point to the base.

Let's begin with the familiar formula for the area A of a triangle:

$\displaystyle A=\frac{1}{2}bh$

Now, using the distance formula, we find:

$\displaystyle b=\sqrt{(x_2-x_1)^2+(y_2-x_y)^2}$

The line through the first two points, using the point-slope formula is:

$\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Arranging this in slope-intercept form, we find:

$\displaystyle y=\frac{y_2-y_1}{x_2-x_1}x-\frac{x_1y_2-y_1x_2}{x_2-x_1}$

Now, using the formula for the distance between a point and a line, we find:

$\displaystyle h=\frac{\left|\frac{y_2-y_1}{x_2-x_1}x_3-\frac{x_1y_2-y_1x_2}{x_2-x_1}-y_3 \right|}{\sqrt{\left(\frac{y_2-y_1}{x_2-x_1} \right)^2+1}}$

$\displaystyle h=\frac{\left|(y_2-y_1)x_3-(x_1y_2-y_1x_2)-y_3(x_2-x_1) \right|}{\sqrt{(x_2-x_1)^2+(y_2-x_y)^2}}$

$\displaystyle h=\frac{\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|}{\sqrt{(x_2-x_1)^2+(y_2-x_y)^2}}$

And so we have:

$\displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|$

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...area-triangle-formed-3-points-plane-4217.html
 
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