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Another way to write equation of the line passing through two points?
up vote
2
down vote
favorite
I am trying to write equation of the line passing through two points pA={1, -3} and pB={-33, -1} in the form x+17 y+50=0. I tried
{pA, pB} = {{1, -3}, {-33, -1}};
u = pB - pA;
m = {x, y};
v = m - pA;
d = Det[{u, v}];
w = {Coefficient[d, x], Coefficient[d, y]};
k = GCD[Coefficient[d, x], Coefficient[d, y]];
If[w[[1]] != 0, n = Sign[w[[1]]] w/k,
If[w[[2]] != 0, n = Sign[w[[2]]] w/k]];
TraditionalForm[Expand[n.v]] == 0
I got
x+17 y+50==0
Is there another way to write it?
output-formatting geometry
asked 15 hours ago
minhthien_2016
544310
|
show 2 more comments
up vote
2
down vote
favorite
I am trying to write equation of the line passing through two points pA={1, -3} and pB={-33, -1} in the form x+17 y+50=0. I tried
{pA, pB} = {{1, -3}, {-33, -1}};
u = pB - pA;
m = {x, y};
v = m - pA;
d = Det[{u, v}];
w = {Coefficient[d, x], Coefficient[d, y]};
k = GCD[Coefficient[d, x], Coefficient[d, y]];
If[w[[1]] != 0, n = Sign[w[[1]]] w/k,
If[w[[2]] != 0, n = Sign[w[[2]]] w/k]];
TraditionalForm[Expand[n.v]] == 0
I got
x+17 y+50==0
Is there another way to write it?
output-formatting geometry
asked 15 hours ago
minhthien_2016
544310
Write or solve?
– Kuba♦
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
You can moveyon the other side of==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.
– Kuba♦
15 hours ago
1
Isn't17 x-y-20==0already in that form?
– Kuba♦
15 hours ago
1
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to write equation of the line passing through two points pA={1, -3} and pB={-33, -1} in the form x+17 y+50=0. I tried
{pA, pB} = {{1, -3}, {-33, -1}};
u = pB - pA;
m = {x, y};
v = m - pA;
d = Det[{u, v}];
w = {Coefficient[d, x], Coefficient[d, y]};
k = GCD[Coefficient[d, x], Coefficient[d, y]];
If[w[[1]] != 0, n = Sign[w[[1]]] w/k,
If[w[[2]] != 0, n = Sign[w[[2]]] w/k]];
TraditionalForm[Expand[n.v]] == 0
I got
x+17 y+50==0
Is there another way to write it?
output-formatting geometry
asked 15 hours ago
minhthien_2016
544310
I am trying to write equation of the line passing through two points pA={1, -3} and pB={-33, -1} in the form x+17 y+50=0. I tried
{pA, pB} = {{1, -3}, {-33, -1}};
u = pB - pA;
m = {x, y};
v = m - pA;
d = Det[{u, v}];
w = {Coefficient[d, x], Coefficient[d, y]};
k = GCD[Coefficient[d, x], Coefficient[d, y]];
If[w[[1]] != 0, n = Sign[w[[1]]] w/k,
If[w[[2]] != 0, n = Sign[w[[2]]] w/k]];
TraditionalForm[Expand[n.v]] == 0
I got
x+17 y+50==0
Is there another way to write it?
output-formatting geometry
output-formatting geometry
asked 15 hours ago
minhthien_2016
544310
asked 15 hours ago
minhthien_2016
544310
edited 1 hour ago
asked 15 hours ago
minhthien_2016
544310
asked 15 hours ago
minhthien_2016
544310
asked 15 hours ago
minhthien_2016
544310
544310
Write or solve?
– Kuba♦
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
You can moveyon the other side of==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.
– Kuba♦
15 hours ago
1
Isn't17 x-y-20==0already in that form?
– Kuba♦
15 hours ago
1
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago
|
show 2 more comments
Write or solve?
– Kuba♦
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
You can moveyon the other side of==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.
– Kuba♦
15 hours ago
1
Isn't17 x-y-20==0already in that form?
– Kuba♦
15 hours ago
1
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago
Write or solve?
– Kuba♦
15 hours ago
Write or solve?
– Kuba♦
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
You can move
y on the other side of ==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.– Kuba♦
15 hours ago
You can move
y on the other side of ==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.– Kuba♦
15 hours ago
1
1
Isn't
17 x-y-20==0 already in that form?– Kuba♦
15 hours ago
Isn't
17 x-y-20==0 already in that form?– Kuba♦
15 hours ago
1
1
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago
|
show 2 more comments
5 Answers
5
active
oldest
votes
up vote
6
down vote
accepted
Simplify[y - InterpolatingPolynomial[{pA, pB}, x] == 0]
50 + x + 17 y == 0
Also
Simplify[y - a x - b == 0 /. First@Solve[a # + b == #2 & @@@ {pA, pB}, {a, b}]]
50 + x + 17 y == 0
And
Simplify @ Rationalize[y - Fit[{pA, pB}, {x, 1}, x] == 0]
50 + x + 17 y == 0
answered 15 hours ago
kglr
171k8194399
add a comment |
up vote
4
down vote
You may do as follows. Let us look for the equation in the form ax+by==1, where the parameters a and b are to be found.This will substitute the coordinates of the points pA and pB into this equations, thus, forming two equations with respect to a and b and solves the system:
eq = a*#[[1]] + b*#[[2]] == 1 &;
eq1=eq /@ {{1, -3}, {-33, -1}}
(* {a - 3 b == 1, -33 a - b == 1} *)
This will substitute the solution into the linear equation already in coordinates x and y:
eq[{x, y}] /. sol
(* -(x/50) - (17 y)/50 == 1 *)
This will plot the solution:
Show[{
ContourPlot[-(x/50) - (17 y)/50 == 1, {x, -34, 2}, {y, -4, 0}],
Graphics[{Red, PointSize[0.015], Point[#] & /@ {{1, -3}, {-33, -1}}}]
}]
yielding the following plot:
The original points are shown in red.
This is one of several possible ways.
Have fun!
answered 15 hours ago
Alexei Boulbitch
21.1k2369
add a comment |
up vote
4
down vote
The equation of a planar line going through two points $ P_1(x_1,y_1) $ and $ P_2(x_2,y_2) $ (e.g. cf this for reference) is
$$
begin{vmatrix}
x & y\
x_2-x_1 & y_2-y_1
end{vmatrix}
=
begin{vmatrix}
x_1 & y_1\
x_2 & y_2
end{vmatrix}.
$$
So there is the piece of codes below:
Clear[eq, pts]
eq = Simplify[Det[{{x, y}, -Subtract @@ #}] == Det[#]] &;
pts = {{1, -3}, {-33, -1}};
eq[pts]
50 + x + 17 y == 0
Or
eq2 = Simplify[Det[{-1, 1} Differences[Prepend[#, {x, y}]]] == 0] &;
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
Ifpts = {{1, -3}, {-33, 150}}How can I get the form9 x+2 y-3=0. Your code ouput9 x+2 y==3Allways in the forma x + b y + c==0,a>0, ifa=0, thenb >0`.
– minhthien_2016
12 hours ago
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
add a comment |
up vote
4
down vote
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered 10 hours ago
halmir
10.1k2443
add a comment |
up vote
2
down vote
Knowing that the coefficients are components of a vector perpendicular to the difference of the two points, I think the most convenient command to obtain the equation is
perp = Cross[pB - pA];
perp.{x, y} == perp.pA // Simplify
50 + x + 17 y == 0
The last step before Simplify is
-2 x - 34 y == 100
so you can see that the simplification brought all terms to one side, factored out the greatest common divisor and fixed the signs.
To Mathematica the sums "50 + x + 17y" and "x + 17y + 50" are exactly the same expression, but if you want to order linear terms before constants in the displayed form, you may consider using TraditionalForm (with the added benefit of using a "normal" equality sign while remaining copy-and-pastable):
% // TraditionalForm
$x+17 y+50=0$
answered 10 hours ago
The Vee
1,393916
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Simplify[y - InterpolatingPolynomial[{pA, pB}, x] == 0]
50 + x + 17 y == 0
Also
Simplify[y - a x - b == 0 /. First@Solve[a # + b == #2 & @@@ {pA, pB}, {a, b}]]
50 + x + 17 y == 0
And
Simplify @ Rationalize[y - Fit[{pA, pB}, {x, 1}, x] == 0]
50 + x + 17 y == 0
answered 15 hours ago
kglr
171k8194399
add a comment |
up vote
6
down vote
accepted
Simplify[y - InterpolatingPolynomial[{pA, pB}, x] == 0]
50 + x + 17 y == 0
Also
Simplify[y - a x - b == 0 /. First@Solve[a # + b == #2 & @@@ {pA, pB}, {a, b}]]
50 + x + 17 y == 0
And
Simplify @ Rationalize[y - Fit[{pA, pB}, {x, 1}, x] == 0]
50 + x + 17 y == 0
answered 15 hours ago
kglr
171k8194399
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Simplify[y - InterpolatingPolynomial[{pA, pB}, x] == 0]
50 + x + 17 y == 0
Also
Simplify[y - a x - b == 0 /. First@Solve[a # + b == #2 & @@@ {pA, pB}, {a, b}]]
50 + x + 17 y == 0
And
Simplify @ Rationalize[y - Fit[{pA, pB}, {x, 1}, x] == 0]
50 + x + 17 y == 0
answered 15 hours ago
kglr
171k8194399
Simplify[y - InterpolatingPolynomial[{pA, pB}, x] == 0]
50 + x + 17 y == 0
Also
Simplify[y - a x - b == 0 /. First@Solve[a # + b == #2 & @@@ {pA, pB}, {a, b}]]
50 + x + 17 y == 0
And
Simplify @ Rationalize[y - Fit[{pA, pB}, {x, 1}, x] == 0]
50 + x + 17 y == 0
answered 15 hours ago
kglr
171k8194399
edited 14 hours ago
answered 15 hours ago
kglr
171k8194399
answered 15 hours ago
kglr
171k8194399
answered 15 hours ago
kglr
171k8194399
171k8194399
add a comment |
add a comment |
up vote
4
down vote
You may do as follows. Let us look for the equation in the form ax+by==1, where the parameters a and b are to be found.This will substitute the coordinates of the points pA and pB into this equations, thus, forming two equations with respect to a and b and solves the system:
eq = a*#[[1]] + b*#[[2]] == 1 &;
eq1=eq /@ {{1, -3}, {-33, -1}}
(* {a - 3 b == 1, -33 a - b == 1} *)
This will substitute the solution into the linear equation already in coordinates x and y:
eq[{x, y}] /. sol
(* -(x/50) - (17 y)/50 == 1 *)
This will plot the solution:
Show[{
ContourPlot[-(x/50) - (17 y)/50 == 1, {x, -34, 2}, {y, -4, 0}],
Graphics[{Red, PointSize[0.015], Point[#] & /@ {{1, -3}, {-33, -1}}}]
}]
yielding the following plot:
The original points are shown in red.
This is one of several possible ways.
Have fun!
answered 15 hours ago
Alexei Boulbitch
21.1k2369
add a comment |
up vote
4
down vote
You may do as follows. Let us look for the equation in the form ax+by==1, where the parameters a and b are to be found.This will substitute the coordinates of the points pA and pB into this equations, thus, forming two equations with respect to a and b and solves the system:
eq = a*#[[1]] + b*#[[2]] == 1 &;
eq1=eq /@ {{1, -3}, {-33, -1}}
(* {a - 3 b == 1, -33 a - b == 1} *)
This will substitute the solution into the linear equation already in coordinates x and y:
eq[{x, y}] /. sol
(* -(x/50) - (17 y)/50 == 1 *)
This will plot the solution:
Show[{
ContourPlot[-(x/50) - (17 y)/50 == 1, {x, -34, 2}, {y, -4, 0}],
Graphics[{Red, PointSize[0.015], Point[#] & /@ {{1, -3}, {-33, -1}}}]
}]
yielding the following plot:
The original points are shown in red.
This is one of several possible ways.
Have fun!
answered 15 hours ago
Alexei Boulbitch
21.1k2369
add a comment |
up vote
4
down vote
up vote
4
down vote
You may do as follows. Let us look for the equation in the form ax+by==1, where the parameters a and b are to be found.This will substitute the coordinates of the points pA and pB into this equations, thus, forming two equations with respect to a and b and solves the system:
eq = a*#[[1]] + b*#[[2]] == 1 &;
eq1=eq /@ {{1, -3}, {-33, -1}}
(* {a - 3 b == 1, -33 a - b == 1} *)
This will substitute the solution into the linear equation already in coordinates x and y:
eq[{x, y}] /. sol
(* -(x/50) - (17 y)/50 == 1 *)
This will plot the solution:
Show[{
ContourPlot[-(x/50) - (17 y)/50 == 1, {x, -34, 2}, {y, -4, 0}],
Graphics[{Red, PointSize[0.015], Point[#] & /@ {{1, -3}, {-33, -1}}}]
}]
yielding the following plot:
The original points are shown in red.
This is one of several possible ways.
Have fun!
answered 15 hours ago
Alexei Boulbitch
21.1k2369
You may do as follows. Let us look for the equation in the form ax+by==1, where the parameters a and b are to be found.This will substitute the coordinates of the points pA and pB into this equations, thus, forming two equations with respect to a and b and solves the system:
eq = a*#[[1]] + b*#[[2]] == 1 &;
eq1=eq /@ {{1, -3}, {-33, -1}}
(* {a - 3 b == 1, -33 a - b == 1} *)
This will substitute the solution into the linear equation already in coordinates x and y:
eq[{x, y}] /. sol
(* -(x/50) - (17 y)/50 == 1 *)
This will plot the solution:
Show[{
ContourPlot[-(x/50) - (17 y)/50 == 1, {x, -34, 2}, {y, -4, 0}],
Graphics[{Red, PointSize[0.015], Point[#] & /@ {{1, -3}, {-33, -1}}}]
}]
yielding the following plot:
The original points are shown in red.
This is one of several possible ways.
Have fun!
answered 15 hours ago
Alexei Boulbitch
21.1k2369
answered 15 hours ago
Alexei Boulbitch
21.1k2369
answered 15 hours ago
Alexei Boulbitch
21.1k2369
answered 15 hours ago
Alexei Boulbitch
21.1k2369
21.1k2369
add a comment |
add a comment |
up vote
4
down vote
The equation of a planar line going through two points $ P_1(x_1,y_1) $ and $ P_2(x_2,y_2) $ (e.g. cf this for reference) is
$$
begin{vmatrix}
x & y\
x_2-x_1 & y_2-y_1
end{vmatrix}
=
begin{vmatrix}
x_1 & y_1\
x_2 & y_2
end{vmatrix}.
$$
So there is the piece of codes below:
Clear[eq, pts]
eq = Simplify[Det[{{x, y}, -Subtract @@ #}] == Det[#]] &;
pts = {{1, -3}, {-33, -1}};
eq[pts]
50 + x + 17 y == 0
Or
eq2 = Simplify[Det[{-1, 1} Differences[Prepend[#, {x, y}]]] == 0] &;
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
Ifpts = {{1, -3}, {-33, 150}}How can I get the form9 x+2 y-3=0. Your code ouput9 x+2 y==3Allways in the forma x + b y + c==0,a>0, ifa=0, thenb >0`.
– minhthien_2016
12 hours ago
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
add a comment |
up vote
4
down vote
The equation of a planar line going through two points $ P_1(x_1,y_1) $ and $ P_2(x_2,y_2) $ (e.g. cf this for reference) is
$$
begin{vmatrix}
x & y\
x_2-x_1 & y_2-y_1
end{vmatrix}
=
begin{vmatrix}
x_1 & y_1\
x_2 & y_2
end{vmatrix}.
$$
So there is the piece of codes below:
Clear[eq, pts]
eq = Simplify[Det[{{x, y}, -Subtract @@ #}] == Det[#]] &;
pts = {{1, -3}, {-33, -1}};
eq[pts]
50 + x + 17 y == 0
Or
eq2 = Simplify[Det[{-1, 1} Differences[Prepend[#, {x, y}]]] == 0] &;
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
Ifpts = {{1, -3}, {-33, 150}}How can I get the form9 x+2 y-3=0. Your code ouput9 x+2 y==3Allways in the forma x + b y + c==0,a>0, ifa=0, thenb >0`.
– minhthien_2016
12 hours ago
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
The equation of a planar line going through two points $ P_1(x_1,y_1) $ and $ P_2(x_2,y_2) $ (e.g. cf this for reference) is
$$
begin{vmatrix}
x & y\
x_2-x_1 & y_2-y_1
end{vmatrix}
=
begin{vmatrix}
x_1 & y_1\
x_2 & y_2
end{vmatrix}.
$$
So there is the piece of codes below:
Clear[eq, pts]
eq = Simplify[Det[{{x, y}, -Subtract @@ #}] == Det[#]] &;
pts = {{1, -3}, {-33, -1}};
eq[pts]
50 + x + 17 y == 0
Or
eq2 = Simplify[Det[{-1, 1} Differences[Prepend[#, {x, y}]]] == 0] &;
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
The equation of a planar line going through two points $ P_1(x_1,y_1) $ and $ P_2(x_2,y_2) $ (e.g. cf this for reference) is
$$
begin{vmatrix}
x & y\
x_2-x_1 & y_2-y_1
end{vmatrix}
=
begin{vmatrix}
x_1 & y_1\
x_2 & y_2
end{vmatrix}.
$$
So there is the piece of codes below:
Clear[eq, pts]
eq = Simplify[Det[{{x, y}, -Subtract @@ #}] == Det[#]] &;
pts = {{1, -3}, {-33, -1}};
eq[pts]
50 + x + 17 y == 0
Or
eq2 = Simplify[Det[{-1, 1} Differences[Prepend[#, {x, y}]]] == 0] &;
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
edited 14 hours ago
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
answered 15 hours ago
Αλέξανδρος Ζεγγ
3,4631927
3,4631927
Ifpts = {{1, -3}, {-33, 150}}How can I get the form9 x+2 y-3=0. Your code ouput9 x+2 y==3Allways in the forma x + b y + c==0,a>0, ifa=0, thenb >0`.
– minhthien_2016
12 hours ago
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
add a comment |
Ifpts = {{1, -3}, {-33, 150}}How can I get the form9 x+2 y-3=0. Your code ouput9 x+2 y==3Allways in the forma x + b y + c==0,a>0, ifa=0, thenb >0`.
– minhthien_2016
12 hours ago
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
If
pts = {{1, -3}, {-33, 150}} How can I get the form 9 x+2 y-3=0 . Your code ouput 9 x+2 y==3 Allways in the form a x + b y + c==0, a>0, if a=0, then b >0`.– minhthien_2016
12 hours ago
If
pts = {{1, -3}, {-33, 150}} How can I get the form 9 x+2 y-3=0 . Your code ouput 9 x+2 y==3 Allways in the form a x + b y + c==0, a>0, if a=0, then b >0`.– minhthien_2016
12 hours ago
@minhthien_2016 Sort of
eq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.– Αλέξανδρος Ζεγγ
11 hours ago
@minhthien_2016 Sort of
eq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.– Αλέξανδρος Ζεγγ
11 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
Thank you very much.
– minhthien_2016
10 hours ago
add a comment |
up vote
4
down vote
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered 10 hours ago
halmir
10.1k2443
add a comment |
up vote
4
down vote
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered 10 hours ago
halmir
10.1k2443
add a comment |
up vote
4
down vote
up vote
4
down vote
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered 10 hours ago
halmir
10.1k2443
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered 10 hours ago
halmir
10.1k2443
answered 10 hours ago
halmir
10.1k2443
answered 10 hours ago
halmir
10.1k2443
answered 10 hours ago
halmir
10.1k2443
10.1k2443
add a comment |
add a comment |
up vote
2
down vote
Knowing that the coefficients are components of a vector perpendicular to the difference of the two points, I think the most convenient command to obtain the equation is
perp = Cross[pB - pA];
perp.{x, y} == perp.pA // Simplify
50 + x + 17 y == 0
The last step before Simplify is
-2 x - 34 y == 100
so you can see that the simplification brought all terms to one side, factored out the greatest common divisor and fixed the signs.
To Mathematica the sums "50 + x + 17y" and "x + 17y + 50" are exactly the same expression, but if you want to order linear terms before constants in the displayed form, you may consider using TraditionalForm (with the added benefit of using a "normal" equality sign while remaining copy-and-pastable):
% // TraditionalForm
$x+17 y+50=0$
answered 10 hours ago
The Vee
1,393916
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
add a comment |
up vote
2
down vote
Knowing that the coefficients are components of a vector perpendicular to the difference of the two points, I think the most convenient command to obtain the equation is
perp = Cross[pB - pA];
perp.{x, y} == perp.pA // Simplify
50 + x + 17 y == 0
The last step before Simplify is
-2 x - 34 y == 100
so you can see that the simplification brought all terms to one side, factored out the greatest common divisor and fixed the signs.
To Mathematica the sums "50 + x + 17y" and "x + 17y + 50" are exactly the same expression, but if you want to order linear terms before constants in the displayed form, you may consider using TraditionalForm (with the added benefit of using a "normal" equality sign while remaining copy-and-pastable):
% // TraditionalForm
$x+17 y+50=0$
answered 10 hours ago
The Vee
1,393916
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Knowing that the coefficients are components of a vector perpendicular to the difference of the two points, I think the most convenient command to obtain the equation is
perp = Cross[pB - pA];
perp.{x, y} == perp.pA // Simplify
50 + x + 17 y == 0
The last step before Simplify is
-2 x - 34 y == 100
so you can see that the simplification brought all terms to one side, factored out the greatest common divisor and fixed the signs.
To Mathematica the sums "50 + x + 17y" and "x + 17y + 50" are exactly the same expression, but if you want to order linear terms before constants in the displayed form, you may consider using TraditionalForm (with the added benefit of using a "normal" equality sign while remaining copy-and-pastable):
% // TraditionalForm
$x+17 y+50=0$
answered 10 hours ago
The Vee
1,393916
Knowing that the coefficients are components of a vector perpendicular to the difference of the two points, I think the most convenient command to obtain the equation is
perp = Cross[pB - pA];
perp.{x, y} == perp.pA // Simplify
50 + x + 17 y == 0
The last step before Simplify is
-2 x - 34 y == 100
so you can see that the simplification brought all terms to one side, factored out the greatest common divisor and fixed the signs.
To Mathematica the sums "50 + x + 17y" and "x + 17y + 50" are exactly the same expression, but if you want to order linear terms before constants in the displayed form, you may consider using TraditionalForm (with the added benefit of using a "normal" equality sign while remaining copy-and-pastable):
% // TraditionalForm
$x+17 y+50=0$
answered 10 hours ago
The Vee
1,393916
answered 10 hours ago
The Vee
1,393916
answered 10 hours ago
The Vee
1,393916
answered 10 hours ago
The Vee
1,393916
1,393916
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
add a comment |
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
Ad sign of the x coefficient: I don't know of any trick simpler than multiply the result by –1 if you don't like it.
– The Vee
10 hours ago
add a comment |
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Write or solve?
– Kuba♦
15 hours ago
@Kuba Write the equation of the line passing through two points.
– minhthien_2016
15 hours ago
You can move
yon the other side of==. This fits it, doesn't it? Or are you asking about different representations of line equation, isn't it a maths question then? Sorry, just trying to get the point.– Kuba♦
15 hours ago
1
Isn't
17 x-y-20==0already in that form?– Kuba♦
15 hours ago
1
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
– Kuba♦
15 hours ago