How to find 3rd coodinates of a triangle if two coordinates and distances between the points are known?
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Assume a simple graph, having a triangle in it. The points are $A, B$ and $C$. I know points $A(x,y)$, $B(x,y)$ and the distances between $A$, $B$ and $C$. I want to get the coordinates of $C(x,y)$.
Eg :- $A(x,y) = A(30,30)$, $B(x,y) = B(90,30)$. The distance between $A$ and $B$ is $60~textcm$, $A$ and $C$ is $50~textcm$ and $B$ and $C$ is $50~textcm$. How to find $C(x,y)$?
geometry triangle coordinate-systems
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
asked Sep 7 at 10:32
muhammed uk
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Assume a simple graph, having a triangle in it. The points are $A, B$ and $C$. I know points $A(x,y)$, $B(x,y)$ and the distances between $A$, $B$ and $C$. I want to get the coordinates of $C(x,y)$.
Eg :- $A(x,y) = A(30,30)$, $B(x,y) = B(90,30)$. The distance between $A$ and $B$ is $60~textcm$, $A$ and $C$ is $50~textcm$ and $B$ and $C$ is $50~textcm$. How to find $C(x,y)$?
geometry triangle coordinate-systems
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
asked Sep 7 at 10:32
muhammed uk
1
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Assume a simple graph, having a triangle in it. The points are $A, B$ and $C$. I know points $A(x,y)$, $B(x,y)$ and the distances between $A$, $B$ and $C$. I want to get the coordinates of $C(x,y)$.
Eg :- $A(x,y) = A(30,30)$, $B(x,y) = B(90,30)$. The distance between $A$ and $B$ is $60~textcm$, $A$ and $C$ is $50~textcm$ and $B$ and $C$ is $50~textcm$. How to find $C(x,y)$?
geometry triangle coordinate-systems
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
asked Sep 7 at 10:32
muhammed uk
1
Assume a simple graph, having a triangle in it. The points are $A, B$ and $C$. I know points $A(x,y)$, $B(x,y)$ and the distances between $A$, $B$ and $C$. I want to get the coordinates of $C(x,y)$.
Eg :- $A(x,y) = A(30,30)$, $B(x,y) = B(90,30)$. The distance between $A$ and $B$ is $60~textcm$, $A$ and $C$ is $50~textcm$ and $B$ and $C$ is $50~textcm$. How to find $C(x,y)$?
geometry triangle coordinate-systems
geometry triangle coordinate-systems
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
asked Sep 7 at 10:32
muhammed uk
1
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
asked Sep 7 at 10:32
muhammed uk
1
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
edited Sep 7 at 10:51
N. F. Taussig
39.7k93153
39.7k93153
asked Sep 7 at 10:32
muhammed uk
1
asked Sep 7 at 10:32
muhammed uk
1
asked Sep 7 at 10:32
muhammed uk
1
1
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52
add a comment |Â
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52
add a comment |Â
1 Answer
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Observe that point $C$ is $50~textcm$ from both points $A$ and $B$.
The locus of points that is $50~textcm$ from $A$ is the circle with radius $50~textcm$ centered at $A$, which is
$$(x - 30~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag1$$
The locus of points that is $50~textcm$ from $B$ is the circle with radius $50~textcm$ centered at $B$, which is
$$(x - 90~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag2$$
Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations
beginalign*
(x - 20)^2 + (y - 20)^2 & = 30^2\
(x - 60)^2 + (y - 20)^2 & = 60^2
endalign*
You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
 |Â
show 10 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Observe that point $C$ is $50~textcm$ from both points $A$ and $B$.
The locus of points that is $50~textcm$ from $A$ is the circle with radius $50~textcm$ centered at $A$, which is
$$(x - 30~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag1$$
The locus of points that is $50~textcm$ from $B$ is the circle with radius $50~textcm$ centered at $B$, which is
$$(x - 90~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag2$$
Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations
beginalign*
(x - 20)^2 + (y - 20)^2 & = 30^2\
(x - 60)^2 + (y - 20)^2 & = 60^2
endalign*
You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
 |Â
show 10 more comments
up vote
0
down vote
Observe that point $C$ is $50~textcm$ from both points $A$ and $B$.
The locus of points that is $50~textcm$ from $A$ is the circle with radius $50~textcm$ centered at $A$, which is
$$(x - 30~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag1$$
The locus of points that is $50~textcm$ from $B$ is the circle with radius $50~textcm$ centered at $B$, which is
$$(x - 90~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag2$$
Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations
beginalign*
(x - 20)^2 + (y - 20)^2 & = 30^2\
(x - 60)^2 + (y - 20)^2 & = 60^2
endalign*
You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
 |Â
show 10 more comments
up vote
0
down vote
up vote
0
down vote
Observe that point $C$ is $50~textcm$ from both points $A$ and $B$.
The locus of points that is $50~textcm$ from $A$ is the circle with radius $50~textcm$ centered at $A$, which is
$$(x - 30~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag1$$
The locus of points that is $50~textcm$ from $B$ is the circle with radius $50~textcm$ centered at $B$, which is
$$(x - 90~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag2$$
Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations
beginalign*
(x - 20)^2 + (y - 20)^2 & = 30^2\
(x - 60)^2 + (y - 20)^2 & = 60^2
endalign*
You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
Observe that point $C$ is $50~textcm$ from both points $A$ and $B$.
The locus of points that is $50~textcm$ from $A$ is the circle with radius $50~textcm$ centered at $A$, which is
$$(x - 30~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag1$$
The locus of points that is $50~textcm$ from $B$ is the circle with radius $50~textcm$ centered at $B$, which is
$$(x - 90~textcm)^2 + (y - 30~textcm)^2 = (50~textcm)^2 tag2$$
Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations
beginalign*
(x - 20)^2 + (y - 20)^2 & = 30^2\
(x - 60)^2 + (y - 20)^2 & = 60^2
endalign*
You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
edited Sep 7 at 13:38
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
answered Sep 7 at 10:49
N. F. Taussig
39.7k93153
39.7k93153
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
 |Â
show 10 more comments
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
Can i apply the same when the values are A(20,20) , B(60,20), and AB=40, BC=60 and AC=20 ?
– muhammed uk
Sep 7 at 11:44
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
It is not possible for a triangle to have side lengths of $20$, $40$, and $60$ since the sum of the lengths of any two sides of a triangle must exceed the length of the third side. If you made $BC = 50$, you could use the same method. The calculations would not be as easy.
– N. F. Taussig
Sep 7 at 11:48
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
Sorry, I typed wrong.
– muhammed uk
Sep 7 at 11:51
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
AB =40, BC =60 and AC =30.
– muhammed uk
Sep 7 at 11:52
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
Yes, you could use the same method I used above. Since $AC = 30$, you would obtain the equation $(x - 20)^2 + (y - 20)^2 = 30^2$. Since $BC = 60$, you would obtain the equation $(x - 60)^2 + (y - 20)^2 = 60^2$. You could then solve for the $x$-coordinate of point $C$ by subtracting the second equation from the first, then use your result to determine the possible values of the $y$-coordinates.
– N. F. Taussig
Sep 7 at 11:59
 |Â
show 10 more comments
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Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
– N. F. Taussig
Sep 7 at 10:52