Solver Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b

Source code of 'Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b'

This Solver (Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b) was created by by hummingbird(0)

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==section input
Find the perpendicular distance between 
point P xo *[input xo=4], yo *[input yo=4] and
line L: y = *[input a=1]x + *[input b=2]

==section solution
perl
my $d=abs($yo-$b);
my $x= $a*$xo+$b;
if( $a == 0 ) {
 print "Coefficients for x is zero, Given <A HREF=Mathematical_line.wikipedia>line</A>line is of the form of {{{y= k}}}(constant) which
 is parallel to X-axis. In this case the required distance will be the difference in the 
 y co-ordinates. Hence, the required distance in this special case is equal to {{{$d}}}\n";
 return;
 }
else {
if( $x== $yo ) {
 print "The given point ($xo,$yo) lies on the given line. Hence, the required distance 
 will be {{{0}}} \n";
 return;
 }
else {
my $x1=$xo;
my $y1= $a*$xo + $b;
my $x2= ($yo-$b)/$a;
my $y2= $yo;
my $PA = sqrt(($x1-$xo)*($x1-$xo) + ($y1-$yo)*($y1-$yo));
my $PB = sqrt(($x2-$xo)*($x2-$xo) + ($y2-$yo)*($y2-$yo));
my $AB = sqrt(($x1-$x2)*($x1-$x2) + ($y1-$y2)*($y1-$y2));
my $height = ($PA*$PB)/$AB;
print

"We want to find the perpendicular distance between a point given by coordinates ({{{$xo}}},{{{$yo}}})
and a line given by equation {{{y=$a*x+$b}}}

First, let's draw a diagram of general situation with point P (xo, yo) and 
line L: y= a.x + b. The required distance is PC. (in the diagram below)

{{{drawing( 200, 200, -2, 8, -2, 8, line(-2,-2,8,8),line(0,0,0,4),line(0,4,4,4),green(line(0,4,2,2)),locate(0,5,P(xo,yo)),locate(0,0,A(x1,y1)), locate(4,4,B(x2,y2)),locate(2,2,C),locate(6,6,L),  red(circle(0,0,.1)),red(circle(4,4,.1)) ) }}}

Methodology
We will first find the vertices of the triangle in order to get the side lengths and then by applying 
<A HREF=Sine_rule.wikipedia>Sine Rule</A> on <A HREF=Special_right_triangles.wikipedia>right angle triangle</A> PAB and PBC we will calculate the desired distance PC.

Step1
Calculation of the vertices of triangle PAB:

Draw a vertical line passing through the point 'P'. This line {{{x=$xo}}} will cut the given line 'L' 
at point 'A'. The X coordinate of A(x1) will be same as {{{xo=$xo}}}. To find the Y-coordinate of
'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation 
of the line 'L' .
Now, plug this {{{x1=$xo}}} in to the equation of line: y=$a*x+$b 
{{{y1=$a*$xo +$b}}}
{{{y1=$y1}}}

Hence, Point (A)({{{x1=$x1}}},{{{y1=$y1}}})

Similarly, 
Draw a horizontal line passing through the point 'P'. This line {{{y=$yo}}} will cut the given line 'L' 
at point 'B'. The Y coordinate of B(y2) will be same as {{{yo=$yo}}}. To find the X-coordinate of
B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation 
of the line 'L' .
Now, plug this {{{y2=$yo}}} in to the equation of line: y=$a*x+$b
{{{$yo=$a*x2+$b}}}
{{{x2= ($yo-$b)/$a}}}
{{{x2=$x2}}}

Hence, Point (B)({{{x2=$x2}}},{{{y2=$y2}}})

Now, we have all the vertices of the triangle PAB

Step2
Calculation of the side lengths using <A HREF=Distance_formula.wikipedia>distance formula</A>:

{{{d=sqrt((x1-x2)^2 + (y1-y2)^2)}}}

Hence, The side lengths PA, PB and AB are
{{{PA=$PA}}}
{{{PB=$PB}}}
{{{AB=$AB}}}

Step3
Apply Sine rule on common angle B in triangle PAB and triangle PBC.
Both triangle PAB and triangle PBC are <A HREF=Special_right_triangles.wikipedia>right angle triangle</A> and points 'A', 'B' and 'C' lay on the given line L.

{{{Sin(B)= AP/AB=PC/BP}}}

{{{PC=(AP*BP)/AB= $height}}}

PC is the required perpendicular distance of the point P ($xo, $yo) from line given 
lineL1: y=$a*x+$b.

For better understanding of this concept, look at the Lesson based on the above concept. 
 <A HREF=http://www.algebra.com/algebra/homework/Coordinate-system/Distance-of-a-Point-From-a-Line.lesson> Lesson </A>";
}
}
==section output
height

==section check
  xo=-1 yo=5 a=1 b=0 height=