math help please!

Question

Show that the triangle with vertices A, B, and C is a right triangle. When the points are A(10,6) B(1,-3) and C(5,-2)

 

[d(A, C)]² =


[d(A, B)]² + [d(B, C)]² =

 

The area is=

Answer 1

This is a test of understanding the Pythagorean theorem, and something closely related, the distance formula in two dimensional cartesian coordinates.

 

We are familiar with the fact that IF a triangle is a right triangle, then a²+b²=c², where c is the length of the hypotenuse (the side opposite the right angle), and a and b are the lengths of the remaining two sides. Another way to put it is to say that c = sqrt{a²+b²): the length of the hypotenuse is equal to the (positive) square root of the sum of the squares of the lengths of the remaining two sides.

 

What we must also be familiar with is that the converse is also true: IF a²+b²=c² THEN a triangle is a right triangle. In other words, when we know the lengths of all three sides of a triangle, we can test whether it is a right triangle by using the pythagorean theorem: first we use the fact that the longest of these sides must be the hypotenuse, if it is a right triangle. Then, if the values match up, the triangle is a right triangle. If they don't match up, it isn't.

 

We use the distance formula to compute the lengths of the sides of the triangle given in this problem. The distance formula is as follows: [d(A,B)]² is equal to the square root of (x₁-x₂)² + (y₁-y₂)², where (x₁,y₁) are the coordinates of A, and (x₂,y₂) are the coordinates of B. In summary:

[d(A,C)]²= square root of (5-10)²+(-2-6)² = 25+64 = 89

[d(A,B)]²= 81 + 81 = 162

[d(B,C)]²= 16 + 1 = 17

The longest side is AB, so if there is a hypotenuse it must be AB. Does 89 + 17 = 162? No. The triangle is not a right triangle.

 

Finally, to compute the area, we must point out that it isn't a right triangle, and therefore we don't know how to do that yet. If it were a right triangle, we could multiply the lengths of the two legs (the two sides other than the hypotenuse) and then divide by two.

 

Further along in a linear algebra course, the area can be found easily by taking half the CROSS PRODUCT of the vectors formed by two of the sides, for example, and likewise the fact it isn't right can be shown by demonstrating that the DOT PRODUCT of the vectors formed by any two potential legs is not 0.