Vertical/Horizontal Distance to a Line

lndist.html

Exploration

Show that the vertical distance, , from a point to a line whose equation is

A x+B y + C=0

is given by

and the horizontal distance, , is given by

.

Approach

Construct a vertical (horizontal) line through the given point. Intersect the vertical (horizontal) line with the given line. The required distance, (), is the distance between and the intersection point.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Construct the vertical (horizontal) line through the given point.

Clear[x1,y1];
p1=Point2D[x1,y1];
{lv=Line2D[p1,Infinity], lh=Line2D[p1,0]}

Intersect the vertical (horizontal) line with the given line.

Clear[a,b,c];
l1=Line2D[a,b,c];
{pv=Point2D[l1,lv],ph=Point2D[l1,lh]};

Find the distance between the intersection point and . The expressions given by Mathematica are equivalent to the desired results.

{Distance2D[p1,pv],Distance2D[p1,ph]}

Discussion

If the point is on the line, then both distances are clearly zero since the point satisfies the equation of the line. If the line has a slope of ±1 (A=±B), then . If the given line is vertical (horizontal), then the vertical (horizontal) distance formula is invalid (i.e. A or B is zero).  Here's a function for vertical distance. The function for horizontal distance would be similar.

Distance2D[Point2D[{x1_,y1_}],
           Line2D[A2_,B2_,C2_],
           Vertical2D]:=
   Abs[(A2*x1+B2*y1+C2)/B2] /;
Not[IsZero2D[B2]]

This computes the vertical distance from (9,2) to the line 2x-4y-3=0.

Distance2D[Point2D[{9,2}],Line2D[2,-4,-3],Vertical2D]