What is the distance from a given line to a given point?

Given a line of the form Ax+By+C=0 and a point (h,k), the perpendicular distance from the line to the point is

|Ah+Bk+C|/sqrt(A²+B²)

Proof

Consider all the lines perpendicular to Ax+By+C=0.  They have the form Bx-Ay+D=0, where A and B are fixed, and D varies to select one of the infinitely many perpendicular lines.  To select the line that goes through (h,k), just select D=-(Bh-Ak), so the equation of the line perpendicular to

Ax+By+C=0

that passes through (h,k) is

Bx-Ay-(Bh-Ak)=0

Solving the two lines to find their point of intersection gives us

x = (B²h-ABk-AC)/(A²+B²)

and

y = (A²k-ABh-BC)/(A²+B²)

Now, the distance from this point of intersection (x,y) to (h,k) is

sqrt((x-h)²+(y-k)²), and then plugging in x and y gives us
sqrt(((B²h-ABk-AC)/(A²+B²)-h)²+((A²k-ABh-BC)/(A²+B²)-k)²),

which, amazingly, simplifies to

sqrt((Ah+Bk+C)²/(A²+B²))

so the final solution is

|Ah+Bk+C|/sqrt(A²+B²)

Internet references

Related pages in this website

Lines in 3 dimensions: Equation of a line in three dimensions, the distance between two lines, finding the closest points of the two lines.

The distance from Ax+By+C=0 to a point is used in Tangents to 2 Circles.

An aside in Point and Triangle2 gives the distance d_AB from (x,y) to line AB where A=(x₁,y₁), B=(x₂,y₂)

Vectors, and their dot product, cross product.

The webmaster and author of this Math Help site is Graeme McRae.