How to find 3rd coodinates of a triangle if two coordinates and distances between the points are known?
Clash Royale CLAN TAG#URR8PPP
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
add a comment |Â
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
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
favorite
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
1
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
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908494%2fhow-to-find-3rd-coodinates-of-a-triangle-if-two-coordinates-and-distances-betwee%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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