2D Coordinate - Plot a line parallel to another line
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.
I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.
How can I do this?
geometry trigonometry coordinate-systems
asked Nov 26 at 8:44
Eric Kim
1073
add a comment |
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.
I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.
How can I do this?
geometry trigonometry coordinate-systems
asked Nov 26 at 8:44
Eric Kim
1073
add a comment |
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.
I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.
How can I do this?
geometry trigonometry coordinate-systems
asked Nov 26 at 8:44
Eric Kim
1073
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.
I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.
How can I do this?
geometry trigonometry coordinate-systems
geometry trigonometry coordinate-systems
asked Nov 26 at 8:44
Eric Kim
1073
asked Nov 26 at 8:44
Eric Kim
1073
asked Nov 26 at 8:44
Eric Kim
1073
asked Nov 26 at 8:44
Eric Kim
1073
asked Nov 26 at 8:44
Eric Kim
1073
1073
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Consider a unitary vector $hat{u}_1$ along the direction of line 1,
$$
hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
$$
The location you are looking for is
$$
mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
$$
answered Nov 26 at 8:50
caverac
13.1k21029
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
add a comment |
First you know x2 and y2 (the coordinates of the beginnning of line 2)
Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.
If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$
All you need to do is move point using the vector (adding them together):
$$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed
Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$
answered Nov 26 at 9:06
TheD0ubleT
39218
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider a unitary vector $hat{u}_1$ along the direction of line 1,
$$
hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
$$
The location you are looking for is
$$
mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
$$
answered Nov 26 at 8:50
caverac
13.1k21029
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
add a comment |
Consider a unitary vector $hat{u}_1$ along the direction of line 1,
$$
hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
$$
The location you are looking for is
$$
mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
$$
answered Nov 26 at 8:50
caverac
13.1k21029
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
add a comment |
Consider a unitary vector $hat{u}_1$ along the direction of line 1,
$$
hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
$$
The location you are looking for is
$$
mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
$$
answered Nov 26 at 8:50
caverac
13.1k21029
Consider a unitary vector $hat{u}_1$ along the direction of line 1,
$$
hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
$$
The location you are looking for is
$$
mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
$$
answered Nov 26 at 8:50
caverac
13.1k21029
answered Nov 26 at 8:50
caverac
13.1k21029
answered Nov 26 at 8:50
caverac
13.1k21029
answered Nov 26 at 8:50
caverac
13.1k21029
13.1k21029
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
add a comment |
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
– Eric Kim
Nov 26 at 8:55
1
1
@EricKim Happy to help
– caverac
Nov 26 at 8:56
@EricKim Happy to help
– caverac
Nov 26 at 8:56
add a comment |
First you know x2 and y2 (the coordinates of the beginnning of line 2)
Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.
If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$
All you need to do is move point using the vector (adding them together):
$$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed
Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$
answered Nov 26 at 9:06
TheD0ubleT
39218
add a comment |
First you know x2 and y2 (the coordinates of the beginnning of line 2)
Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.
If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$
All you need to do is move point using the vector (adding them together):
$$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed
Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$
answered Nov 26 at 9:06
TheD0ubleT
39218
add a comment |
First you know x2 and y2 (the coordinates of the beginnning of line 2)
Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.
If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$
All you need to do is move point using the vector (adding them together):
$$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed
Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$
answered Nov 26 at 9:06
TheD0ubleT
39218
First you know x2 and y2 (the coordinates of the beginnning of line 2)
Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.
If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$
All you need to do is move point using the vector (adding them together):
$$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed
Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$
answered Nov 26 at 9:06
TheD0ubleT
39218
answered Nov 26 at 9:06
TheD0ubleT
39218
answered Nov 26 at 9:06
TheD0ubleT
39218
answered Nov 26 at 9:06
TheD0ubleT
39218
39218
add a comment |
add a comment |
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