Doug C. answered • 03/26/20
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The coordinates of the centroid of a triangle are found by finding the intersection of the equations for two of the medians. A median joins a vertex with the midpoint of the opposite side. Determine two of the midpoints. Find the slope of the median, then use point-slope to write the equation. Ditto for the 2nd median. Post a comment here is you need more detail.
Doug C.
Yes there is an easier way. The centroid is 2/3 of the distance from the vertex to the midpoint. This is like a problem that asks "find the coordinate of the point on a segment that divides the segment into pieces that are in the ratio 2:1".03/26/20
Drew M.
I am now struggling on finding the base and the height of this triangle. If its an equilateral then i can split it in half and use the Pythagorean theorem. But what does c= so that i can find b. a=203/27/20
Doug C.
A sketch should reveal what is going on. Should see an isosceles triangle with base from (1,2) to (1,6) and y = 4 passes through (7,4) and the midpoint of the base. So the segment joining (1,4) and (7,4) is an altitude and a median. The centroid must be 2/3 of the directed distance from x=7 to x=1 along the line y = 4. The base is a vertical line and the altitude is horizontal. Attach this to the url for Desmos to see this is true: /calculator/dduatmjzbg03/27/20
Doug C.
To check your final answer see the following: desmos.com/calculator/u8fhlc6exk03/27/20
Drew M.
I saw the desmos example and thanks for the all of your advice. For what I got was similar to the calculators answer. I got 1/2(b*h) which was 1/2(4*6) =12. 2piR was 2pi times ybar which gave me 8pi. 8pi times 12 gave me 96pi as my final answer.03/28/20
Drew M.
Also I can't use a graphing calculator if I have this problem on a test. How do i find the base and the height of triangles without graphing? Somewhere on google look like y2-y1 and x3-x2 idk its confusing do you know the formula for this?03/28/20
Drew M.
I notice when I know all the 3 vertices of a triangle the following approach works to find the base and the height and works everytime. X3-X1 gives me the height and Y2-Y1 gives me the base. Works everytime03/28/20
Doug C.
Finding base and height the way you suggest is only going to work if a side and altitude are horizontal and vertical (of vice-versa). That was the case with this problem. The real key is how would you find the area of a triangle knowing the coordinates of 3 vertices in general? Do a search for "shoelace formula/algorithm". Another option would be to use the distance formula to find the length of each side then use Heron's formula--complicated. My guess on a test you will likely have a base that is horizontal or vertical. Determine that by noticing that two vertices have the same x-coordinate or same y-coordinate. The altitude from the other vertex will be a perpendicular segment and its length easy to calculate. In your problem two of the vertices have the same x-coordinate (1,2) and (1,6). So that side is vertical with a length of 6 - 2. The altitude from (7,4) will be perpendicular to the base. The length of that altitude will be 7 - 1, change in x. Change to coordinates of (1,6) to (3,9), for example. Finding the area will not be so easy.03/28/20
Drew M.
Can this be done without graphing? I feel that there is a more efficient way to do this because finding the slope of the medians this way seems quite lengthy when people in my class skip this step entirely. I'll just refer back to the textbook. Thank you Doug!03/26/20