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Mathematics 14 Online
OpenStudy (anonymous):

done/got help

OpenStudy (acxbox22):

you have to find the slope of all three segments

OpenStudy (anonymous):

@acxbox22 what do you mean? also please read my question again i redid something i did wrong the first time.

jimthompson5910 (jim_thompson5910):

you made an error with the first slope

jimthompson5910 (jim_thompson5910):

the slope formula is m = (y2-y1)/(x2-x1) the y difference is on top, the x difference is down below

jimthompson5910 (jim_thompson5910):

but you mixed up x and y for that first slope

OpenStudy (anonymous):

Yeah I know and Ohhh I see thanks Jim give me a sec to redo!

OpenStudy (acxbox22):

well use the formula above to find the slope of all three segments made by the three points then we examine the slopes out of the three slopes if two are perpendicular slopes than it is a right triangle

jimthompson5910 (jim_thompson5910):

oh wait, nvm, I misread the only error you made on the first one is that the slope is actually +0.2 and not -0.2

jimthompson5910 (jim_thompson5910):

in any event, the two slopes don't multiply to -1 0.2*(-4) = -0.8 so the two lines are NOT perpendicular

OpenStudy (anonymous):

So it is not a right triangle? So I did that part correctly?

jimthompson5910 (jim_thompson5910):

you need to show that the other pairs of segments aren't perpendicular

OpenStudy (acxbox22):

yes

OpenStudy (anonymous):

other pair of segments? so I need to find the slope between (2,2) and (6,7)?

jimthompson5910 (jim_thompson5910):

X(2,2) Y(6,7) Z(7,3) you found the slope of line YZ to be -4 you found the slope of line XZ to be 0.2

jimthompson5910 (jim_thompson5910):

find the slope of XY and use it to see if XY is perpendicular to either YZ or XZ

OpenStudy (anonymous):

Ahhh okay I see now! give me a sec okay?

jimthompson5910 (jim_thompson5910):

sure

OpenStudy (anonymous):

oops ignore the first K idk why its their lol

jimthompson5910 (jim_thompson5910):

slope of line YZ is -4 slope of line XZ is 0.2 slope of line XY is 1.25

jimthompson5910 (jim_thompson5910):

let m1 = slope of YZ m2 = slope of XZ m3 = slope of XY if m1*m2 = -1 or m1*m3 = -1 or m2*m3 = -1 if any of the above statements are true, then we have a pair of perpendicular lines

jimthompson5910 (jim_thompson5910):

looks good

jimthompson5910 (jim_thompson5910):

none of those results are -1, so we don't have a right triangle

OpenStudy (anonymous):

Btw thanks for having patience with me and making this easy :] Alright so now I have to figure out what changes could be made to make it a right triangle.

jimthompson5910 (jim_thompson5910):

hmm let me think

OpenStudy (anonymous):

Could changing the coordinates work?

jimthompson5910 (jim_thompson5910):

yes it turns out that all you need to do is move 1 point

jimthompson5910 (jim_thompson5910):

there are infinitely many ways to do this, but one way is to move point Y up one unit. Recompute the slopes and confirm that you do indeed get a pair of perpendicular lines.

jimthompson5910 (jim_thompson5910):

X(2,2) Y(6,7) ---> turns into Y(6,8) Z(7,3) compute the slopes of XY, XZ, YZ

jimthompson5910 (jim_thompson5910):

you made an error here YZ = 8-4/6-7 = 4/-1 = - 4

jimthompson5910 (jim_thompson5910):

I don't know where you got the 4 in "8-4/6-7"

jimthompson5910 (jim_thompson5910):

I made X and Z stay where they are. I only moved Y

jimthompson5910 (jim_thompson5910):

anything involving point Z

jimthompson5910 (jim_thompson5910):

multiply those slopes and see which multiply to -1

jimthompson5910 (jim_thompson5910):

so XZ and YZ are perpendicular, making a right angle occur at their intersection point

OpenStudy (anonymous):

yay awesome! Is this it? Did we really work out the whole problem? lol sorry my head actually hurts from all this but I am truly thankful for your help it was amazing and you were very patient and made me understand this which is beyond amazing! Thank you.

jimthompson5910 (jim_thompson5910):

yeah that's the full thing and I'm glad it's all making sense now

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