Im trying to find the 3rd set of coordinates when I know the other two coordinates and I know the lengths of 2 sides - example: If I had a right triangle "a,b,c" and I know angle b is 90°, I know the length of ab is .89, I know the length of bc is .5, I know the coordinates of a are (1,7) and I know the coordinates of b are (2,3) how do I solve the coordinates of c?. thank you.
need help finding 3rd set of coordinates to a right triangle
- Thread starter cja7928
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What have you tried? We need to see your work to help where you are stuck.
Something in the problem is wrong: the distance between the given points a and b is NOT 0.89.
Do you know how to use slopes to make a line perpendicular to ab, through b?
I wasnt really sure where to start. .89 may not be correct I just sketched my example on some graph paper and it looked to be a hair under .9 No I havnt learned about slopes. is that a good place to start? thank you for your help.
D
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Do you know how to calculate the distance between two points whose co-ordinates are (x1,y1) & (x2,y2)?
If you don't - that will be the first place to start.
If you do - calculate the distance between points A and B.
so I came up with
√(2-1)² + (3-7)²
√(1)² + (-4)²
√1 + 16
√17
4.12
does this look correct?
thank you!
cja7928, that appears as if you just threw together an example that doesn't work.
Correct.... I am an idiot. I did present a bad example. .89 is way off, not sure what I was thinking. It's been a long day.
ab = 4.12 bc = .5
I appreciate all the help. What is the next step to find the x,y of c?
Thanks again for your input.
I probably need more than classroom help.....
Yes .89 is way off It should have been 4.12
i apologize. Thanks for your time.
I would be interested in learning how to calculate the 3rd missing set if coordinates of a right triangle using whichever example is best. The example I presented represents the triangle I need to calculate for but if there is a better way to learn I am very interested. Thank you
D
Deleted member 4993
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If you are given co-ordinates of only two points (vertex) of a right angled triangle - there infinite sets points for the third vertex.
You need to post a real problem - and you need tell us the context of the problem (e.g. carpentry work, home-work from class-room) along with your capabilities (like which level of math you finished or attending).
its a right triangle abc.
point a coordinates = (1,7)
point b coordinates = (2,3)
length ab = 4.12
length bc = .5
angle b is 90°
is it not possible to calculate the coordinates for point c from the given information?
D
Deleted member 4993
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Yes - it is possible, but the answer is not unique. There are two sets of solutions.
But before doing that - I need to know,
Its a long story - its not related to carpentry or home work. Im trying to calculate the position 1/2" up perpendicular to an angled surface so I know where to locate a piece of equipment automatically. I know the coordinates of 2 points on that angled surface. Im affraid trying to adequately explain the situation might be confusing and lead us in circles. My math education ended at high school geometry. I unfortunately was not able to acquire a further education past high school. everything ive done thus far has been self taught but I have been unable to find the solution for this particular problem. perhaps my problem is i dont know what to search for - should I be looking for slope rather than a right triangle? Thank you for your time.
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D
Deleted member 4993
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Okay so then answers are: (2.485071,3.12168) & (1.514929, 2.878732)
It is a slightly complicated problem involving simple geometry and algebra and "nasty" arithmetic.
You'll round-off the answer, depending on how close to 0.500 you want the height(BC) to be (and how close to 90° you want that angle ABC to be).
Suppose you take those coordinates to be (2.5, 3.1) - then your height will be BC = 0.51 and the angle ABC = 87.27°.
Wow, thank you. Can you show me how you did it. I would like to learn. Thank you
Yea I did the same search, just had a hard time following those examples. First one starts out with Pn which I wasn't really sure what that was I didn't see an n in the example.
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Here's another method, using similarity. Consider the following diagram:
We know then that:
\(\displaystyle \dfrac{\left|y-y_1 \right|}{m}=\dfrac{\left|x_1-x_2 \right|}{n}\)
\(\displaystyle \left|y-y_1 \right|=\dfrac{m}{n}\left|x_1-x_2 \right|\)
\(\displaystyle y-y_1=\pm\dfrac{m}{n}\left(x_1-x_2 \right)\)
\(\displaystyle y=y_1\pm\dfrac{m}{n}\left(x_1-x_2 \right)\)
and:
\(\displaystyle \dfrac{\left|x-x_1 \right|}{m}=\dfrac{\left|y_2-y_1 \right|}{n}\)
\(\displaystyle \left|x-x_1 \right|=\dfrac{m}{n}\left|y_2-y_1 \right|\)
\(\displaystyle x-x_1=\pm\dfrac{m}{n}\left(y_2-y_1 \right)\)
\(\displaystyle x=x_1\pm\dfrac{m}{n}\left(y_2-y_1 \right)\)
And so we have:
\(\displaystyle (x,y)=\left(x_1\pm\dfrac{m\left(y_2-y_1 \right)}{n},y_1\pm\dfrac{m\left(x_1-x_2 \right)}{n} \right)\)
We know then that:
\(\displaystyle \dfrac{\left|y-y_1 \right|}{m}=\dfrac{\left|x_1-x_2 \right|}{n}\)
\(\displaystyle \left|y-y_1 \right|=\dfrac{m}{n}\left|x_1-x_2 \right|\)
\(\displaystyle y-y_1=\pm\dfrac{m}{n}\left(x_1-x_2 \right)\)
\(\displaystyle y=y_1\pm\dfrac{m}{n}\left(x_1-x_2 \right)\)
and:
\(\displaystyle \dfrac{\left|x-x_1 \right|}{m}=\dfrac{\left|y_2-y_1 \right|}{n}\)
\(\displaystyle \left|x-x_1 \right|=\dfrac{m}{n}\left|y_2-y_1 \right|\)
\(\displaystyle x-x_1=\pm\dfrac{m}{n}\left(y_2-y_1 \right)\)
\(\displaystyle x=x_1\pm\dfrac{m}{n}\left(y_2-y_1 \right)\)
And so we have:
\(\displaystyle (x,y)=\left(x_1\pm\dfrac{m\left(y_2-y_1 \right)}{n},y_1\pm\dfrac{m\left(x_1-x_2 \right)}{n} \right)\)
D
Deleted member 4993
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That's a very elegant solution Mark.
I did it by "brute" force - equating distances.
I did it by "brute" force - equating distances.
D
Deleted member 4993
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Denis,
OP had defined the co-ordinates of A and C. We need to find the co-ordinates of B in that system. Your simplified answer (in the rotated co-ordinate system) needs to be rotated back to the original system.
Lookagain - it is not lookagain!!
OP had defined the co-ordinates of A and C. We need to find the co-ordinates of B in that system. Your simplified answer (in the rotated co-ordinate system) needs to be rotated back to the original system.
Lookagain - it is not lookagain!!
Wow, thats exactly what I needed. thank you! I appreciate the breakdown that really helps. very nice. thank you.