Another way to write equation of the line passing through two points? [closed]
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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 Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
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closed as unclear what you're asking by march, José Antonio Díaz Navas, m_goldberg, Bob Hanlon, Sumit Nov 22 '18 at 19:46
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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 Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
$endgroup$
closed as unclear what you're asking by march, José Antonio Díaz Navas, m_goldberg, Bob Hanlon, Sumit Nov 22 '18 at 19:46
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
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@Kuba Write the equation of the line passing through two points.
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– minhthien_2016
Nov 19 '18 at 8:43
1
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Isn't17 x-y-20==0already in that form?
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– Kuba♦
Nov 19 '18 at 8:51
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Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
1
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58
add a comment |
$begingroup$
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 Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
$endgroup$
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 Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
asked Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
edited Nov 19 '18 at 23:09
minhthien_2016
asked Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
asked Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
asked Nov 19 '18 at 8:37
minhthien_2016minhthien_2016
566310
566310
closed as unclear what you're asking by march, José Antonio Díaz Navas, m_goldberg, Bob Hanlon, Sumit Nov 22 '18 at 19:46
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by march, José Antonio Díaz Navas, m_goldberg, Bob Hanlon, Sumit Nov 22 '18 at 19:46
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
$begingroup$
@Kuba Write the equation of the line passing through two points.
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:43
1
$begingroup$
Isn't17 x-y-20==0already in that form?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:51
$begingroup$
Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
1
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58
add a comment |
$begingroup$
Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
$begingroup$
@Kuba Write the equation of the line passing through two points.
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:43
1
$begingroup$
Isn't17 x-y-20==0already in that form?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:51
$begingroup$
Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
1
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58
$begingroup$
Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
$begingroup$
Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
$begingroup$
@Kuba Write the equation of the line passing through two points.
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:43
$begingroup$
@Kuba Write the equation of the line passing through two points.
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:43
1
1
$begingroup$
Isn't
17 x-y-20==0 already in that form?$endgroup$
– Kuba♦
Nov 19 '18 at 8:51
$begingroup$
Isn't
17 x-y-20==0 already in that form?$endgroup$
– Kuba♦
Nov 19 '18 at 8:51
$begingroup$
Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
$begingroup$
Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
1
1
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58
add a comment |
6 Answers
6
active
oldest
votes
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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 Nov 19 '18 at 9:03
kglrkglr
183k10202417
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add a comment |
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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 Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
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add a comment |
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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 Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
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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`.
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– minhthien_2016
Nov 19 '18 at 12:18
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@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
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– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
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Thank you very much.
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– minhthien_2016
Nov 19 '18 at 13:46
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Another way :Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify
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– Sigis K
Nov 20 '18 at 23:16
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@SigisK Yes, it is.
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– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
add a comment |
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With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
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add a comment |
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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 Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
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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.
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– The Vee
Nov 19 '18 at 13:26
add a comment |
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Although this might be more a math question, the ingenious answers have taught me a lot about MMA. Thanks to all contributors. I can add an answer based on my high school Analytical Geometry classes. The mnemonic two-point form of the equation of a straight line through A(x1,y1) and B(x2,y2) is
Using the coordinates for pA and pB given in the question:
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
y2 - y1) == (x - x1)/(x2 - x1)] // TraditionalForm
gives
However the more 'symmetrical form' of the equation leads to an answer that is mathematically identical, but is not in the required form. Does anybody know how to force MMA to yield the required form?
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
x - x1) == (y2 - y1)/(x2 - x1)] // TraditionalForm
gives
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
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I think, we can write in the formSimplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm
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– minhthien_2016
Nov 21 '18 at 6:29
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My way is almost write the equations in the forma x + b y + c = 0.
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– minhthien_2016
Nov 21 '18 at 6:48
add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Nov 19 '18 at 9:03
kglrkglr
183k10202417
$endgroup$
add a comment |
$begingroup$
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 Nov 19 '18 at 9:03
kglrkglr
183k10202417
$endgroup$
add a comment |
$begingroup$
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 Nov 19 '18 at 9:03
kglrkglr
183k10202417
$endgroup$
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 Nov 19 '18 at 9:03
kglrkglr
183k10202417
edited Nov 19 '18 at 9:32
answered Nov 19 '18 at 9:03
kglrkglr
183k10202417
answered Nov 19 '18 at 9:03
kglrkglr
183k10202417
answered Nov 19 '18 at 9:03
kglrkglr
183k10202417
183k10202417
add a comment |
add a comment |
$begingroup$
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 Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
$endgroup$
add a comment |
$begingroup$
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 Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
$endgroup$
add a comment |
$begingroup$
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 Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
$endgroup$
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 Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
answered Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
answered Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
answered Nov 19 '18 at 9:17
Alexei BoulbitchAlexei Boulbitch
21.5k2470
21.5k2470
add a comment |
add a comment |
$begingroup$
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 Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
$endgroup$
$begingroup$
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`.
$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Another way :Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify
$endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
add a comment |
$begingroup$
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 Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
$endgroup$
$begingroup$
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`.
$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Another way :Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify
$endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
add a comment |
$begingroup$
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 Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
$endgroup$
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 Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
edited Nov 19 '18 at 9:38
answered Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
answered Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
answered Nov 19 '18 at 9:05
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,2341929
4,2341929
$begingroup$
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`.
$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Another way :Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify
$endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
add a comment |
$begingroup$
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`.
$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
@minhthien_2016 Sort ofeq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Another way :Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify
$endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
$begingroup$
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`.$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
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`.$endgroup$
– minhthien_2016
Nov 19 '18 at 12:18
$begingroup$
@minhthien_2016 Sort of
eq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
@minhthien_2016 Sort of
eq[pts] /. a_ == b_ :> a - b == 0, though I think it just a minor issue.$endgroup$
– Αλέξανδρος Ζεγγ
Nov 19 '18 at 13:04
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Thank you very much.
$endgroup$
– minhthien_2016
Nov 19 '18 at 13:46
$begingroup$
Another way :
Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify $endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
Another way :
Det[{{x - #1[[1]], y - #1[[2]]}, {x - #2[[1]], y - #2[[2]]}}] == 0 & @@ {{1, -3}, {-33, -1}} // Simplify $endgroup$
– Sigis K
Nov 20 '18 at 23:16
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
$begingroup$
@SigisK Yes, it is.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 21 '18 at 4:12
add a comment |
$begingroup$
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
$endgroup$
add a comment |
$begingroup$
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
$endgroup$
add a comment |
$begingroup$
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
$endgroup$
With RegionMember:
Simplify[RegionMember[InfiniteLine[{{1, -3}, {-33, -1}}], {x, y}],
Element[x | y, Reals]]
50 + x + 17 y == 0
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
answered Nov 19 '18 at 13:59
halmirhalmir
10.2k2443
10.2k2443
add a comment |
add a comment |
$begingroup$
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 Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
$endgroup$
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
add a comment |
$begingroup$
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 Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
$endgroup$
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
add a comment |
$begingroup$
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 Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
$endgroup$
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 Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
answered Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
answered Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
answered Nov 19 '18 at 13:25
The VeeThe Vee
1,393916
1,393916
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
add a comment |
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
$begingroup$
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.
$endgroup$
– The Vee
Nov 19 '18 at 13:26
add a comment |
$begingroup$
Although this might be more a math question, the ingenious answers have taught me a lot about MMA. Thanks to all contributors. I can add an answer based on my high school Analytical Geometry classes. The mnemonic two-point form of the equation of a straight line through A(x1,y1) and B(x2,y2) is
Using the coordinates for pA and pB given in the question:
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
y2 - y1) == (x - x1)/(x2 - x1)] // TraditionalForm
gives
However the more 'symmetrical form' of the equation leads to an answer that is mathematically identical, but is not in the required form. Does anybody know how to force MMA to yield the required form?
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
x - x1) == (y2 - y1)/(x2 - x1)] // TraditionalForm
gives
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
$endgroup$
$begingroup$
I think, we can write in the formSimplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
My way is almost write the equations in the forma x + b y + c = 0.
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
add a comment |
$begingroup$
Although this might be more a math question, the ingenious answers have taught me a lot about MMA. Thanks to all contributors. I can add an answer based on my high school Analytical Geometry classes. The mnemonic two-point form of the equation of a straight line through A(x1,y1) and B(x2,y2) is
Using the coordinates for pA and pB given in the question:
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
y2 - y1) == (x - x1)/(x2 - x1)] // TraditionalForm
gives
However the more 'symmetrical form' of the equation leads to an answer that is mathematically identical, but is not in the required form. Does anybody know how to force MMA to yield the required form?
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
x - x1) == (y2 - y1)/(x2 - x1)] // TraditionalForm
gives
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
$endgroup$
$begingroup$
I think, we can write in the formSimplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
My way is almost write the equations in the forma x + b y + c = 0.
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
add a comment |
$begingroup$
Although this might be more a math question, the ingenious answers have taught me a lot about MMA. Thanks to all contributors. I can add an answer based on my high school Analytical Geometry classes. The mnemonic two-point form of the equation of a straight line through A(x1,y1) and B(x2,y2) is
Using the coordinates for pA and pB given in the question:
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
y2 - y1) == (x - x1)/(x2 - x1)] // TraditionalForm
gives
However the more 'symmetrical form' of the equation leads to an answer that is mathematically identical, but is not in the required form. Does anybody know how to force MMA to yield the required form?
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
x - x1) == (y2 - y1)/(x2 - x1)] // TraditionalForm
gives
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
$endgroup$
Although this might be more a math question, the ingenious answers have taught me a lot about MMA. Thanks to all contributors. I can add an answer based on my high school Analytical Geometry classes. The mnemonic two-point form of the equation of a straight line through A(x1,y1) and B(x2,y2) is
Using the coordinates for pA and pB given in the question:
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
y2 - y1) == (x - x1)/(x2 - x1)] // TraditionalForm
gives
However the more 'symmetrical form' of the equation leads to an answer that is mathematically identical, but is not in the required form. Does anybody know how to force MMA to yield the required form?
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)/(
x - x1) == (y2 - y1)/(x2 - x1)] // TraditionalForm
gives
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
answered Nov 20 '18 at 11:57
GommaireGommaire
1664
1664
$begingroup$
I think, we can write in the formSimplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
My way is almost write the equations in the forma x + b y + c = 0.
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
add a comment |
$begingroup$
I think, we can write in the formSimplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
My way is almost write the equations in the forma x + b y + c = 0.
$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
$begingroup$
I think, we can write in the form
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
I think, we can write in the form
Simplify@With[{x1 = 1, y1 = -3, x2 = -33, y2 = -1}, (y - y1)* (x2 - x1) - (x - x1)*(y2 - y1) == 0] // TraditionalForm$endgroup$
– minhthien_2016
Nov 21 '18 at 6:29
$begingroup$
My way is almost write the equations in the form
a x + b y + c = 0.$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
$begingroup$
My way is almost write the equations in the form
a x + b y + c = 0.$endgroup$
– minhthien_2016
Nov 21 '18 at 6:48
add a comment |
$begingroup$
Write or solve?
$endgroup$
– Kuba♦
Nov 19 '18 at 8:39
$begingroup$
@Kuba Write the equation of the line passing through two points.
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:43
1
$begingroup$
Isn't
17 x-y-20==0already in that form?$endgroup$
– Kuba♦
Nov 19 '18 at 8:51
$begingroup$
Yes. My question is "is there another way to write the equation in that form?"
$endgroup$
– minhthien_2016
Nov 19 '18 at 8:53
1
$begingroup$
You can multiply sides by a constant but I fail to see how it is a Mathematica question.
$endgroup$
– Kuba♦
Nov 19 '18 at 8:58