Determine if triangle XYZ with coordinates X (2, 2), Y (6, 7), and Z (7, 3) is a right triangle. Use evidence to support your claim. If it is not a right triangle, what changes can be made to make it a right triangle? Be specific.
|dw:1417027437221:dw| There are 2 ways I can suggest: First work out the length of all the sides (using the coordinates of the endpoints). Then call the 2 shorter sides a & b , and the other one c if a^2 +b^2 = c^2 then it is a right angle (by Pythagoras) second - Work out the slope of XZ (m1) Work out the slope of YZ (m2) if the they are perpendicular then m1- -1/m2
cani show you what i got?
We first must use the distance formula to find out if the distance fits in to the Pythagorean theorem. Distance formula: Sqrt{(x2 - x1)^2 + (y2 - y1)^2} so we now plug in the coordinates to find distance of X to Y. Sqrt{(6 - 2)^2 + (7-2)^2} Sqrt{(4)^2 + (5)^2} Sqrt{16 + 25} Sqrt{41} Now find distance for Y to Z Sqrt{(7 - 6)^2 + (3 - 7)^2} Sqrt{(1)^2 + (-4)^2} Sqrt{1+ -16} Sqrt{-15} Now find distance for X to Z Sqrt{(7 - 2)^2 + (3 - 2)^2} Sqrt{(5)^2 + (1)^2} Sqrt{25 + 1} Sqrt{26} So now we add 26 and -15 to see if it adds up to XY distance. 26 - 15 = 11 So since 11 is not equal 41, then this triangle is not a right triangle.
I meant m1=-1/m2
and to fix it, all that is needed is make the 3 in point Z a -3
@hartnn
you have made a mistake with your lengths (YZ) Your method is fine, but the calculation is wrong
Sqrt{(7 - 6)^2 + (3 - 7)^2} Sqrt{(1)^2 + (-4)^2} Sqrt{1+ -16} <--here? Sqrt{-15}
should it be 16 not -16
That is not correct Yes -4^2 = 16
yes - you are correct now
ah ok so Now find distance for Y to Z Sqrt{(7 - 6)^2 + (3 - 7)^2} Sqrt{(1)^2 + (-4)^2} Sqrt{1+16} Sqrt{17} 17 + 26 still does not equal 41
correct
ah thanks, what about the second part of the question?
soz - gtg
soz?
ok thanks bye
the easy solution is find the slope of the 3 segments. Multiply the slopes of each pair. If the answer from the multiplication is -1 then the lines are perpendicular.
ah ok that is easier
can you help me with the second part of the question? If it is not a right triangle, what changes can be made to make it a right triangle? Be specific.
for the adjustment, take the slope of a line segment. then for the other segment adjust an endpoint so the the slope is the negative reciprocal of the slope you started with.
ok but i would have to restart right? since i didn't do the slope way
@campbell_st
nvm i got it
thanks to the both of you
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