Question

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Answer 1

1. Show that this is a right triangle, in which case the circumcentre is at the mid-point of the hypotenuse.
2. see 1.
3. By the sine law, and the fact that sin(x)/x is a decreasing function, we conclude that the angles can be ranked by the length of the opposite sides.

Answer 2

This is a long process, I'll give you the formulas you will need.

Answer 3

Circumcenter:

(1) find the midpoint of each side

Midpoint Formula:\[(\frac{x_2+x_1}{2}),(\frac{y_2+y_1}{2})\]

Answer 4

do i that for all of the numbers in parenthesis?

Answer 5

For a right triangle (as is the case here), the circumcentre is the mid-point of the hypotenuse.

I think chml is attempting to show you the general case.
Reply if this is the case.

Answer 6

i cant figure this out for my life i just wanted the formulas and stuff so i could work the answer ot myself

Answer 7

I'm sorry something came up and I had to jump off.

Answer 8

The circumcenter is the point that is equidistant from each vertex

Answer 9

To find this you want the perpendicular bisectors of each side and then fin where they intersect.

Answer 10

1) Find the slope between two sets of points. This will be used to find the equation of the perpendicular bisector.\[\text{slope}=\frac{ y_2-y_1 }{ x_2-x_1 }\]
2) Now that we have the slope, we know that the slope of the perpendicular bisector line will be the opposite reciprocal of that.

For example, if the slope was 1/2 the opposite reciprocal would be -2/1 or just -2. We will use this slope in the point-slope formula for constructing an equation of a line.\[y-y_1=m(x-x_1)\]
3) By now you have noticed that we do not have a point. We will use the midpoint between two of the vertices. To find this use the midpoint formula\[(\frac{x_2+x_1}{2}),(\frac{y_2+y_1}{2})\]
4) Once we have the point, we just simply plug that into (x1,y1) of the point slope form equation, where m is the slope, and we get our equation. You have to do this entire process twice because we need two perpendicular bisecting equations. So just choose 2 sets of points.

Where the perpendicular bisecting equations intersect is the circumcenter.

5) Set the equations equal to eachother and solve for x. Plug this value back into one of the equations to get y and you have your circumcenter.

Answer 11

Question 3) the angle opposite the longest side is the longest and the pattern continues down to the smallest side

Answer 12

@ElizabethScarlette
It is very important that you know the general method of finding circumcentres of any given triangle with known coordinates of vertices, as ChmE had taken the time to show you.

If this method is foreign to you, it is possible that you have not yet done it in class (but perhaps in the following week). Both your problems deal with the special case where the triangles are right triangles. This suggests that you have finished the module that deals with this particular case.
In this special case, you could still use the general method, or you could show:
1. that the triangle is a right triangle (oriented with the x- and y- axes)
This happens when at least one pair of x values are equal, and one pair of y-values are equal. Example, (1,4),(2,4),(1,3) since 1 is the x-value of two vertices, and 4 is the value of two other vertices, the triangle is a right triangle.
2. locate the hypotenuse
The right-angled vertex is the one with common x and y-coordinates, namely (1,4).
So the hypotenuse is the side joining (2,4) and (1,3).
3. calculate the circumcentre:
It is the mid-point of the hypotenuse, using
ChmE's formula, ((2+1)/2, (4+3)/2) = (3/2, 7/2).
4. Finally, if the triangle is not oriented along the axes, it is simpler to use the general method, even though it is possible to verify perpendicularity.