**Notes for area of a triangle defines by three points and the distance formula.**

**Notes for interior angles sums of polygons and the measure of regular angles.**

**More practice on the interior angles sums and regular angle measurements.**

**Classifying triangles defined by three vertices in the plane.**

In class, we have worked with three points in the plane where one is the origin (0, 0) and the other two are labeled (

*a, b*) and (

*c, d*). In the notes here on the blog, the points are given coordinates with subscripts such as (

*x*1,

*y*1) and (

*x*2,

*y*2). The formulas are the same, only the letters used in the formulas are different.

The area for the triangle define by (0, 0), (

*a, b*) and (

*c, d*) is given by the formula

Area = ½|

*ad - bc*|

This formula is relatively simple compared to Heron's formula, with no square roots necessary. On the other hand, we will need to do three square roots to get the lengths of the three line segments for both classification methods. Let's look at an example.

**Example # 1:**(0, 0), (6, 3), (7, 1)

**Area:**½|

*(6)(1) - (3)(7)| = ½| 6 - 21| = ½| - 15 | = ½(15) =*

**7.5****Distances:**

Between (0, 0) and (6, 3): When (0, 0) is involved the subtractions are easy and we get

Distance = sqrt((0-6)² + (0-3)²) = sqrt(6² + 3²) = sqrt(36 + 9) =

**sqrt(45)**Usually, I'd ask to have this simplified, but this problem just leave the answer as the square root of 45, it will make the next steps easier.

Between (0, 0) and (7, 1)

Distance = sqrt((0-7)² + (0-1)²) = sqrt(7² + 1²) = sqrt(49 + 1) =

**sqrt(50)**Between (6, 3) and (7, 1)

Distance = sqrt((6-7)² + (3-1)²) = sqrt((-1)² + 2²) = sqrt(1 + 4) =

**sqrt(5)**Now we have the distances, the classifications can be done.

Equilateral, isosceles or scalene? The numbers are all different sqrt(45), sqrt(50) and sqrt(5), so the triangle is scalene.

Obtuse, right or acute? Here we need to square the distances, which is easy because they are all written as square roots, so the squares are 45, 50 and 5. We need to compare the sum of the two short sides squared versus the long side squared and in this case

__45 + 5 = 50.__

This is a right triangle.

**Final answer:**Area of 7.5 square units, a scalene right triangle.

**Example #2:**(3, 2), (4, 1), (1, 7)

Here, we don't have a point at the origin, so we will subtract a value of one given vertex from all of the vertices. That will give us three new points, but the triangle defined by our new points will have the same area and classifications. I will choose to subtract (3, 2), but using (4, 1) or (1, 7) would also work.

New vertex #1: (3, 2) - (3, 2) = (0, 0)

New vertex #2: (4, 1) - (3, 2) = (1, -1)

New vertex #4: (1, 7) - (3, 2) = (-2, 5)

**Area**= ½|

*(1)(5) - (-1)(-2)| = ½| 5 - 2| = ½| 3 | = ½(3) =*

**1.5**Between (0, 0) and (1, -1):

Distance = sqrt((0-1)² + (0-(-1))²) = sqrt(1² + 1²) = sqrt(1 + 1) =

**sqrt(2)**Usually, I'd ask to have this simplified, but this problem just leave the answer as the square root of 45, it will make the next steps easier.

Between (0, 0) and (-2, 5)

Distance = sqrt((0-(-2))² + (0-5)²) = sqrt(2² + 5²) = sqrt(4 + 25) =

**sqrt(29)**Between (1, -1) and (-2, 5)

Distance = sqrt((1-(-2))² + ((-1)-5)²) = sqrt(3² + (-6)²) = sqrt(9 + 36) =

**sqrt(45)**Now we have the distances, the classifications can be done.

Equilateral, isosceles or scalene? The numbers are all different sqrt(2), sqrt(29) and sqrt(45), so the triangle is scalene.

Obtuse, right or acute? Here we need to square the distances, which is easy because they are all written as square roots, so the squares are 2, 29 and 45. We need to compare the sum of the two short sides squared versus the long side squared and in this case

__2 + 29 = 31 < 45.__

This is an obtuse triangle.

**Final answer:**Area of 1.5 square units, a scalene obtuse triangle.

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