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Calculate third point of triangle from two points and angles
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I've got two points(p1 and p2) and two angles(angle1 and angle2), I can calculate the third angle, but how do I calculate the coordinates of point p? Not just the distances from the points, but coordinates.
I'm trying to use this to do texture mapping on triangles. Here is an image of my idea
p1 = (2, 0)
p2 = (6, 4)
angle1 is angle next to p1,
angle2 is angle next to p2.
trigonometry
asked Apr 3 '16 at 10:34
ca1ek
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bumped to the homepage by Community♦ 20 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
1
down vote
favorite
I've got two points(p1 and p2) and two angles(angle1 and angle2), I can calculate the third angle, but how do I calculate the coordinates of point p? Not just the distances from the points, but coordinates.
I'm trying to use this to do texture mapping on triangles. Here is an image of my idea
p1 = (2, 0)
p2 = (6, 4)
angle1 is angle next to p1,
angle2 is angle next to p2.
trigonometry
asked Apr 3 '16 at 10:34
ca1ek
63
bumped to the homepage by Community♦ 20 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've got two points(p1 and p2) and two angles(angle1 and angle2), I can calculate the third angle, but how do I calculate the coordinates of point p? Not just the distances from the points, but coordinates.
I'm trying to use this to do texture mapping on triangles. Here is an image of my idea
p1 = (2, 0)
p2 = (6, 4)
angle1 is angle next to p1,
angle2 is angle next to p2.
trigonometry
asked Apr 3 '16 at 10:34
ca1ek
63
I've got two points(p1 and p2) and two angles(angle1 and angle2), I can calculate the third angle, but how do I calculate the coordinates of point p? Not just the distances from the points, but coordinates.
I'm trying to use this to do texture mapping on triangles. Here is an image of my idea
p1 = (2, 0)
p2 = (6, 4)
angle1 is angle next to p1,
angle2 is angle next to p2.
trigonometry
trigonometry
asked Apr 3 '16 at 10:34
ca1ek
63
asked Apr 3 '16 at 10:34
ca1ek
63
edited Apr 3 '16 at 11:08
asked Apr 3 '16 at 10:34
ca1ek
63
asked Apr 3 '16 at 10:34
ca1ek
63
asked Apr 3 '16 at 10:34
ca1ek
63
63
bumped to the homepage by Community♦ 20 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 20 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10
add a comment |
1
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10
1
1
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10
add a comment |
3 Answers
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Hint: write the equations linking $(x,y)$ coordinates for points on the green and blue lines. Then solve as a system of simultaneous equations.
Also: you need to clarify if the angles are signed or not, and if the lines are half lines as on the drawing. With full lines and non-signed angles, there are as much as 4 solutions.
answered Apr 3 '16 at 10:57
fgrieu
541319
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Let us fix some notations:
let $alpha_1,alpha_2,alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=sqrt{u^2+v^2}$.
First of all: $alpha_3=pi-(alpha_1+alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$dfrac{a_1}{sin alpha_1}=dfrac{a_2}{sin alpha_2}=dfrac{a_3}{sin alpha_3}$$
one obtains in particular $a_2=a_3dfrac{sin alpha_2}{sin alpha_3}$ where where $alpha_2,alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $vec{p_1p_2}.vec{p_1p_3}=a_2 a_3 cos alpha_1$ and
the norm of the cross product $|vec{p_1p_2}timesvec{p_1p_3}|=a_2 a_3 sin alpha_1$
by using coordinates:
$$begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 cos alpha_1\
u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 sin alpha_1end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v+a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
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I don't understand any of your pictures. Ignoring them, first, find the slope of $overline{P_1P_2}$:
$$m_0 = frac{y_2 - y_1}{x_2 - x_1} = 1$$
The angle associated with that is $arctan(1) = frac{pi}{4}$
Next, find the slope of the lines:
$$m_1 = arctan(frac{pi}{4} - angle_1) , m_2 = arctan(angle_1 - frac{pi}{4}) $$
You may have to be a little intelligent about subtracting those angles. Next, write the lines in point-slope form:
$$y - y_1 = m_1 ( x - x_1) \
y - y_2 = m_2 ( x-x_2)$$
Finally, the point in question is located where those two lines cross:
$$ m_1 ( x - x_1) + y_1 =m_2 ( x - x_2) + y_2 \
x = frac{y_2 - y_1 - m_2 x_2 + m_1 x_1}{ m_1 - m_2} \
y = frac{frac{y_1}{m_1} - frac{y_2}{m_2} + x_2 - x_1}{frac{1}{m_1} - frac{1}{m_2}}$$
The $hat x$ equation is stable unless the slopes of those two lines are very close, so leave that to the computer. The $hat y$ equation, conversely, is unstable, so instead calculate:
$$y = m_1 ( x_p - x_1) + y_1$$
answered Mar 8 at 5:52
user121330
64049
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: write the equations linking $(x,y)$ coordinates for points on the green and blue lines. Then solve as a system of simultaneous equations.
Also: you need to clarify if the angles are signed or not, and if the lines are half lines as on the drawing. With full lines and non-signed angles, there are as much as 4 solutions.
answered Apr 3 '16 at 10:57
fgrieu
541319
add a comment |
up vote
0
down vote
Hint: write the equations linking $(x,y)$ coordinates for points on the green and blue lines. Then solve as a system of simultaneous equations.
Also: you need to clarify if the angles are signed or not, and if the lines are half lines as on the drawing. With full lines and non-signed angles, there are as much as 4 solutions.
answered Apr 3 '16 at 10:57
fgrieu
541319
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: write the equations linking $(x,y)$ coordinates for points on the green and blue lines. Then solve as a system of simultaneous equations.
Also: you need to clarify if the angles are signed or not, and if the lines are half lines as on the drawing. With full lines and non-signed angles, there are as much as 4 solutions.
answered Apr 3 '16 at 10:57
fgrieu
541319
Hint: write the equations linking $(x,y)$ coordinates for points on the green and blue lines. Then solve as a system of simultaneous equations.
Also: you need to clarify if the angles are signed or not, and if the lines are half lines as on the drawing. With full lines and non-signed angles, there are as much as 4 solutions.
answered Apr 3 '16 at 10:57
fgrieu
541319
answered Apr 3 '16 at 10:57
fgrieu
541319
answered Apr 3 '16 at 10:57
fgrieu
541319
answered Apr 3 '16 at 10:57
fgrieu
541319
541319
add a comment |
add a comment |
up vote
0
down vote
Let us fix some notations:
let $alpha_1,alpha_2,alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=sqrt{u^2+v^2}$.
First of all: $alpha_3=pi-(alpha_1+alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$dfrac{a_1}{sin alpha_1}=dfrac{a_2}{sin alpha_2}=dfrac{a_3}{sin alpha_3}$$
one obtains in particular $a_2=a_3dfrac{sin alpha_2}{sin alpha_3}$ where where $alpha_2,alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $vec{p_1p_2}.vec{p_1p_3}=a_2 a_3 cos alpha_1$ and
the norm of the cross product $|vec{p_1p_2}timesvec{p_1p_3}|=a_2 a_3 sin alpha_1$
by using coordinates:
$$begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 cos alpha_1\
u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 sin alpha_1end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v+a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
add a comment |
up vote
0
down vote
Let us fix some notations:
let $alpha_1,alpha_2,alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=sqrt{u^2+v^2}$.
First of all: $alpha_3=pi-(alpha_1+alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$dfrac{a_1}{sin alpha_1}=dfrac{a_2}{sin alpha_2}=dfrac{a_3}{sin alpha_3}$$
one obtains in particular $a_2=a_3dfrac{sin alpha_2}{sin alpha_3}$ where where $alpha_2,alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $vec{p_1p_2}.vec{p_1p_3}=a_2 a_3 cos alpha_1$ and
the norm of the cross product $|vec{p_1p_2}timesvec{p_1p_3}|=a_2 a_3 sin alpha_1$
by using coordinates:
$$begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 cos alpha_1\
u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 sin alpha_1end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v+a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
add a comment |
up vote
0
down vote
up vote
0
down vote
Let us fix some notations:
let $alpha_1,alpha_2,alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=sqrt{u^2+v^2}$.
First of all: $alpha_3=pi-(alpha_1+alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$dfrac{a_1}{sin alpha_1}=dfrac{a_2}{sin alpha_2}=dfrac{a_3}{sin alpha_3}$$
one obtains in particular $a_2=a_3dfrac{sin alpha_2}{sin alpha_3}$ where where $alpha_2,alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $vec{p_1p_2}.vec{p_1p_3}=a_2 a_3 cos alpha_1$ and
the norm of the cross product $|vec{p_1p_2}timesvec{p_1p_3}|=a_2 a_3 sin alpha_1$
by using coordinates:
$$begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 cos alpha_1\
u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 sin alpha_1end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v+a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
Let us fix some notations:
let $alpha_1,alpha_2,alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=sqrt{u^2+v^2}$.
First of all: $alpha_3=pi-(alpha_1+alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$dfrac{a_1}{sin alpha_1}=dfrac{a_2}{sin alpha_2}=dfrac{a_3}{sin alpha_3}$$
one obtains in particular $a_2=a_3dfrac{sin alpha_2}{sin alpha_3}$ where where $alpha_2,alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $vec{p_1p_2}.vec{p_1p_3}=a_2 a_3 cos alpha_1$ and
the norm of the cross product $|vec{p_1p_2}timesvec{p_1p_3}|=a_2 a_3 sin alpha_1$
by using coordinates:
$$begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 cos alpha_1\
u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 sin alpha_1end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v+a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
edited Apr 3 '16 at 14:11
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
answered Apr 3 '16 at 10:43
Jean Marie
28.1k41848
28.1k41848
add a comment |
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I don't understand any of your pictures. Ignoring them, first, find the slope of $overline{P_1P_2}$:
$$m_0 = frac{y_2 - y_1}{x_2 - x_1} = 1$$
The angle associated with that is $arctan(1) = frac{pi}{4}$
Next, find the slope of the lines:
$$m_1 = arctan(frac{pi}{4} - angle_1) , m_2 = arctan(angle_1 - frac{pi}{4}) $$
You may have to be a little intelligent about subtracting those angles. Next, write the lines in point-slope form:
$$y - y_1 = m_1 ( x - x_1) \
y - y_2 = m_2 ( x-x_2)$$
Finally, the point in question is located where those two lines cross:
$$ m_1 ( x - x_1) + y_1 =m_2 ( x - x_2) + y_2 \
x = frac{y_2 - y_1 - m_2 x_2 + m_1 x_1}{ m_1 - m_2} \
y = frac{frac{y_1}{m_1} - frac{y_2}{m_2} + x_2 - x_1}{frac{1}{m_1} - frac{1}{m_2}}$$
The $hat x$ equation is stable unless the slopes of those two lines are very close, so leave that to the computer. The $hat y$ equation, conversely, is unstable, so instead calculate:
$$y = m_1 ( x_p - x_1) + y_1$$
answered Mar 8 at 5:52
user121330
64049
add a comment |
up vote
0
down vote
I don't understand any of your pictures. Ignoring them, first, find the slope of $overline{P_1P_2}$:
$$m_0 = frac{y_2 - y_1}{x_2 - x_1} = 1$$
The angle associated with that is $arctan(1) = frac{pi}{4}$
Next, find the slope of the lines:
$$m_1 = arctan(frac{pi}{4} - angle_1) , m_2 = arctan(angle_1 - frac{pi}{4}) $$
You may have to be a little intelligent about subtracting those angles. Next, write the lines in point-slope form:
$$y - y_1 = m_1 ( x - x_1) \
y - y_2 = m_2 ( x-x_2)$$
Finally, the point in question is located where those two lines cross:
$$ m_1 ( x - x_1) + y_1 =m_2 ( x - x_2) + y_2 \
x = frac{y_2 - y_1 - m_2 x_2 + m_1 x_1}{ m_1 - m_2} \
y = frac{frac{y_1}{m_1} - frac{y_2}{m_2} + x_2 - x_1}{frac{1}{m_1} - frac{1}{m_2}}$$
The $hat x$ equation is stable unless the slopes of those two lines are very close, so leave that to the computer. The $hat y$ equation, conversely, is unstable, so instead calculate:
$$y = m_1 ( x_p - x_1) + y_1$$
answered Mar 8 at 5:52
user121330
64049
add a comment |
up vote
0
down vote
up vote
0
down vote
I don't understand any of your pictures. Ignoring them, first, find the slope of $overline{P_1P_2}$:
$$m_0 = frac{y_2 - y_1}{x_2 - x_1} = 1$$
The angle associated with that is $arctan(1) = frac{pi}{4}$
Next, find the slope of the lines:
$$m_1 = arctan(frac{pi}{4} - angle_1) , m_2 = arctan(angle_1 - frac{pi}{4}) $$
You may have to be a little intelligent about subtracting those angles. Next, write the lines in point-slope form:
$$y - y_1 = m_1 ( x - x_1) \
y - y_2 = m_2 ( x-x_2)$$
Finally, the point in question is located where those two lines cross:
$$ m_1 ( x - x_1) + y_1 =m_2 ( x - x_2) + y_2 \
x = frac{y_2 - y_1 - m_2 x_2 + m_1 x_1}{ m_1 - m_2} \
y = frac{frac{y_1}{m_1} - frac{y_2}{m_2} + x_2 - x_1}{frac{1}{m_1} - frac{1}{m_2}}$$
The $hat x$ equation is stable unless the slopes of those two lines are very close, so leave that to the computer. The $hat y$ equation, conversely, is unstable, so instead calculate:
$$y = m_1 ( x_p - x_1) + y_1$$
answered Mar 8 at 5:52
user121330
64049
I don't understand any of your pictures. Ignoring them, first, find the slope of $overline{P_1P_2}$:
$$m_0 = frac{y_2 - y_1}{x_2 - x_1} = 1$$
The angle associated with that is $arctan(1) = frac{pi}{4}$
Next, find the slope of the lines:
$$m_1 = arctan(frac{pi}{4} - angle_1) , m_2 = arctan(angle_1 - frac{pi}{4}) $$
You may have to be a little intelligent about subtracting those angles. Next, write the lines in point-slope form:
$$y - y_1 = m_1 ( x - x_1) \
y - y_2 = m_2 ( x-x_2)$$
Finally, the point in question is located where those two lines cross:
$$ m_1 ( x - x_1) + y_1 =m_2 ( x - x_2) + y_2 \
x = frac{y_2 - y_1 - m_2 x_2 + m_1 x_1}{ m_1 - m_2} \
y = frac{frac{y_1}{m_1} - frac{y_2}{m_2} + x_2 - x_1}{frac{1}{m_1} - frac{1}{m_2}}$$
The $hat x$ equation is stable unless the slopes of those two lines are very close, so leave that to the computer. The $hat y$ equation, conversely, is unstable, so instead calculate:
$$y = m_1 ( x_p - x_1) + y_1$$
answered Mar 8 at 5:52
user121330
64049
answered Mar 8 at 5:52
user121330
64049
answered Mar 8 at 5:52
user121330
64049
answered Mar 8 at 5:52
user121330
64049
64049
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1
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
– Inazuma
Apr 3 '16 at 10:50
There is obviously not a single, but 2 answers for point $p_3$ (symmetrical with respect to line $p_1p_2$)
– Jean Marie
Apr 3 '16 at 10:53
@JeanMarie I know, I also know how to determine the point I want.
– ca1ek
Apr 3 '16 at 11:10