Question 969431: Find the perimeter of the triangle whose vertices are the following specified points in the plane.
(2,7), (5,-10), and (1,-5)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! You use the distance formula for each of the three points. You can go in either direction, and negative values get squared. BUT, you have to do the ys and the xs separately, not mix them. Unlike slope, you don't have to go in the same direction, but it is better if you do.
From (2,7) to (5,-10), the distance is sqrt [(-10-7)^2 + (5-2)^2]=sqrt (-17^2 + 3^2)=sqrt (289+9)=sqrt (298).
Stated another way, the x changes 3, which we square (9), and the y changes -17,which we square to get 289. We add the two and take the square root.
From (5,-10) to (1,-5) the distance is sqrt [(-5-(-10))^2+(1-5)^2)]=sqrt(5^2+-4^2)=sqrt (25+16)=sqrt (41)
We see that the x changes 4 (decrease or increase doesn't matter, since the square will make it positive. The square is 16. The y changes 5, the square 25. Take the square root of the sum (41)
From (1,-5) to (2,7) the distance is sqrt[(7-(-5))^2 + (2-1)^2}=sqrt (12^2+1^2)=sqrt (145)
We go only 1 unit with the x (square is 1) and 12 units with the y (square 144). We add the two and take their square root (of 145)
sqrt(298)+sqrt(41)+sqrt(145)=17.26+6.40+12.04=35.70 units.
Note: Do NOT add the square roots and then take the square root of the sum. sqrt(4)+sqrt(9)+sqrt(12)=2+3+2 sqrt(3)=5 +2 sqrt(3) NOT sqrt(25)=5, a nicer answer, but not the right one. [In this problem the sqrt(484) is also a "nice" answer (22), but it is not correct.]
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