How can I deduce the distance formula from a point to a straight line?

Question

Consider:

Clear["Global`*"]
reg = ImplicitRegion[a x + b y + c == 0, {x, y}]
pt = {x0, y0}
RegionDistance[reg, pt]

The distance formula from a point to a straight line cannot be derived using the above code.

Without knowing how to solve it, I attempted to replace it with the following code through other means:

Clear["Global`*"]
line=a x+b y+c==0
pt={x0,y0}
pol=Apply[Subtract,line];
Flatten@CoefficientList[pol,{x,y}];
A1=Coefficient[pol,x];
B1=Coefficient[pol,y];
C1=Select[pol,FreeQ[x|y]];
distance=Abs[A1 pt[[1]]+B1 pt[[2]]+C1]/Sqrt[A1^2+B1^2]

How can I change the first code to achieve the goal?

Answer 1

If we check the examples in document of RegionDistance, we'll see most of the examples are about specific graphic primitives (Line, Disk, etc.). This isn't too surprising, because functions like ImplicitRegion are so general, and general problems are always hard to handle. This inspires us to try replacing the ImplicitRegion with an InfiniteLine as a workaround:

ptfunc[y_] = {x /. Solve[a x + b y + c == 0, {x}][[1]], y}
(* {(-c - b y)/a, y} *)

reg = InfiniteLine[ptfunc /@ {0, 1}]
(* InfiniteLine[{{-(c/a), 0}, {(-b - c)/a, 1}}] *)

pt = {x0, y0};
dis = RegionDistance[reg, pt]
(* Sqrt[(c + a x0 + b y0)^2/(a^2 + b^2)] *)

dis // FullSimplify[#, {a, b, c, x0, y0} ∈ Reals] &
(* Abs[c + a x0 + b y0]/Sqrt[a^2 + b^2] *)

---

Update

As pointed out by user64494 in comment below, I've implicitly assumed a!=0 in the solution above, so let me add a rigorous and simpler solution making use of 2nd syntax of Hyperplane:

reg = Hyperplane[{a, b}, -c];

pt = {x0, y0};
dis = RegionDistance[reg, pt]
(* Abs[(c + a x0 + b y0)/Sqrt[a^2 + b^2]] *)

dis // Simplify[#, {a, b} ∈ Reals] &
(* Abs[c + a x0 + b y0]/Sqrt[a^2 + b^2] *)

Answer 2

Another workaround:

First a visualization of a selected instance:

Derivation:

pt = {x0, y0};
line = a x + b y + c == 0;
tri = Triangle[{pt
   , {0, y} /. First@Solve[line, y] /. {x -> 0}
   , {x, 0} /. First@Solve[line, x] /. {y -> 0}
   }]

Triangle[{{x0, y0}, {0, -(c/b)}, {-(c/a), 0}}]

ans = (TriangleMeasurement[tri, {"Height", pt}]) // 
  FullSimplify[#, {a, b, c, x0, y0} ∈ Reals] & (* `FullSimplify` step copied from xzczd's answer*)

Abs[c + a x0 + b y0]/Sqrt[a^2 + b^2]

which after (unnecessary) formatting becomes:

ans /. {x0 -> Subscript[x, 0], y0 -> Subscript[y, 0]} // 
 Map@TraditionalForm

Answer 3

Given a point and a line:

pt={x0,y0}
line= a x+b y+c==0;

If we project the point: pt={x,y} perpendicularly onto the line we have the point: pp on the line closest to pt, giving the distance of the point from the line. Therefore, we have 2 equations one that pp={xp,yp} is on the line. The second comes from: pt-p0 is perpendicular to the direction of the line. Therefor, the scalar product from a vector in direction of the line and (pt-pp) must be zero

pt={x,y};
pp= {xp,yp};

The direction of the line: A vector perpendicular to the line is: {a,b}. Therefore, the direction is: {-b,a} (check: take the scalar product {a,b}.{-b,a})

dir= {-b,a}

Therefore we have 2 equations to determine pp:

eq = {{a, b} . pp + c == 0, dir. (pp - pt) == 0 }

And we get pp by:

ppsol= pp /. Solve[eq, {xp, yp}];

With this we get the distance:

Norm[pt -ppsol]

We can put all together in a distance function:

dist[pt_, {a_, b_, c_}] := 
 Module[{pp = {xp, yp}, dir = {-b, a}, eq, ppsol}, 
  eq = {{a, b} . pp + c == 0, dir . (pt - pp) == 0};
  ppsol = pp /. Solve[eq, pp][[1]]; 
  Norm[pt - ppsol]
  ]

Now, let us try if this works. A line from {0,1} to {1,0} has {a,b,c}=={1,1,-2} and a point {1,1} has a distance of 2/Sqrt[2] from this line:

dist[{2, 2}, {1, 1, -2}]

1/Sqrt[2]

Answer 4

Here is a direct approach which demonstrates the difficulties caused by 5 parameters $a,b,c,x0,y0$.

Minimize[{Sqrt[({x, y} - {x0, y0}) . ({x, y} - {x0, y0})], 
a*x + b*y + c == 0 && a^2 + b^2 > 0}, {x, y}] // First

Piecewise[{{Sqrt[x0^2], (x0 == 0 && c == 0 && a > 0 && b == 0) || (x0 == 0 && c == 0 && a < 0 && b == 0)}, {Sqrt[(c + a*x0)^2/a^2], (x0 > 0 && a > 0 && b == 0) || (x0 > 0 && a < 0 && b == 0) || (x0 < 0 && a > 0 && b == 0) || (x0 < 0 && a < 0 && b == 0)}, {Sqrt[(c^2 + a^2*x0^2)/a^2], (x0 == 0 && c > 0 && a > 0 && b == 0) || (x0 == 0 && c > 0 && a < 0 && b == 0) || (x0 == 0 && c < 0 && a > 0 && b == 0) || (x0 == 0 && c < 0 && a < 0 && b == 0)}, {Sqrt[(c + b*y0)^2/b^2], (x0 > 0 && a == 0 && b > 0) || (x0 > 0 && a == 0 && b < 0) || (x0 < 0 && a == 0 && b > 0) || (x0 < 0 && a == 0 && b < 0)}, {Sqrt[(c + a*x0 + b*y0)^2/(a^2 + b^2)], (x0 < 0 && a < 0 && b > 0) || (x0 > 0 && a > 0 && b > 0) || (x0 > 0 && a > 0 && b < 0) || (x0 > 0 && a < 0 && b > 0) || (x0 > 0 && a < 0 && b < 0) || (x0 < 0 && a > 0 && b > 0) || (x0 < 0 && a > 0 && b < 0) || (x0 < 0 && a < 0 && b < 0)}, {Sqrt[x0^2 + y0^2], (x0 == 0 && c == 0 && a == 0 && b > 0) || (x0 == 0 && c == 0 && a == 0 && b < 0)}, {Sqrt[(c^2 + a^2*x0^2 + b^2*x0^2 + 2*b*c*y0 + b^2*y0^2)/(a^2 + b^2)], (x0 == 0 && c > 0 && a > 0 && b > 0) || (x0 == 0 && c > 0 && a > 0 && b < 0) || (x0 == 0 && c > 0 && a < 0 && b > 0) || (x0 == 0 && c > 0 && a < 0 && b < 0) || (x0 == 0 && c < 0 && a > 0 && b > 0) || (x0 == 0 && c < 0 && a > 0 && b < 0) || (x0 == 0 && c < 0 && a < 0 && b > 0) || (x0 == 0 && c < 0 && a < 0 && b < 0)}, {Sqrt[x0^2 + (b^2*y0^2)/(a^2 + b^2)], (x0 == 0 && c == 0 && a > 0 && b > 0) || (x0 == 0 && c == 0 && a > 0 && b < 0) || (x0 == 0 && c == 0 && a < 0 && b > 0) || (x0 == 0 && c == 0 && a < 0 && b < 0)}, {Sqrt[x0^2 + y0^2 + (c*(c + 2*b*y0))/b^2], (x0 == 0 && c > 0 && a == 0 && b > 0) || (x0 == 0 && c > 0 && a == 0 && b < 0) || (x0 == 0 && c < 0 && a == 0 && b > 0) || (x0 == 0 && c < 0 && a == 0 && b < 0)}}, Infinity]

Its execution takes more than dozen minutes.

FullSimplify[%, Assumptions -> a^2 + b^2 > 0 && {a, b, x0, y0} ∈ Reals]

Piecewise[{{0, b == 0 && c == 0 && x0 == 0}, {Sqrt[(c + a*x0)^2/a^2], b == 0 && x0 != 0}, {Sqrt[c^2/a^2], b == 0 && x0 == 0 && c != 0}, {Sqrt[(c + b*y0)^2/b^2], a == 0 && (c != 0 || x0 != 0)}, {Sqrt[(c + a*x0 + b*y0)^2/(a^2 + b^2)], a != 0 && b != 0 && x0 != 0}, {Abs[y0], a == 0 && c == 0 && x0 == 0}, {Sqrt[(c + b*y0)^2/(a^2 + b^2)], x0 == 0 && a != 0 && b != 0 && c != 0}}, (Abs[b]*Abs[y0])/Sqrt[a^2 + b^2]]

At last,

FullSimplify[Piecewise[{{0, b == 0 && c == 0 && x0 == 0}, {Sqrt[(c + a*x0)^2/a^2], 
    b == 0 && x0 != 0}, {Sqrt[c^2/a^2], 
 b == 0 && x0 == 0 && c != 0}, 
  {Sqrt[(c + b*y0)^2/b^2], a == 0 && (c != 0 || x0 != 0)}, 
  {Sqrt[(c + a*x0 + b*y0)^2/(a^2 + b^2)], 
 a != 0 && b != 0 && x0 != 0}, 
  {Abs[y0], a == 0 && c == 0 && x0 == 0}, 
  {Sqrt[(c + b*y0)^2/(a^2 + b^2)], 
 x0 == 0 && a != 0 && b != 0 && 
      c != 0}}, (Abs[b]*Abs[y0])/Sqrt[a^2 + b^2]] - 
  Sqrt[(c + a*x0 + b*y0)^2/(a^2 + b^2)], 
 Assumptions -> a^2 + b^2 > 0 && {a, b, x0, y0} ∈ Reals]

0

FullSimplify[Sqrt[(c + a*x0 + b*y0)^2/(a^2 + b^2)] - 
RealAbs[(c + a*x0 + b*y0)]/Sqrt[(a^2 + b^2)], 
 Assumptions -> a^2 + b^2 > 0 && {a, b, x0, y0} ∈ Reals]

0