Let the equation of line L be Ax+By+C=0, and the coordinate of point P be (x0, y0), then the distance from point P to line L is:

Similarly, when the analytical formula of P(x0, y0) and straight line L is y=kx+b, the distance from point P to straight line L is:

Consider the point (x0, y0, z0) and the spatial straight line x-x1/l = y-Y 1- 0//m = z-z1/n, where d = | (X 1-x0, y/kloc). +m？ +n？ )。

Proof method:

Definition forensics: According to the definition, point P(x? ，y？ ) Distance to straight line L: ax+by+c = 0 is the vertical length from point P to straight line L. Let the vertical line from point P to straight line L be L' and the vertical foot be Q, then the slope of L' is B/A, and the analytical formula of L' is y-y? =(B/A)(x-x？ The coordinates of the intersection q of l and l' are ((B^2x? Abby? -AC)/(A^2+B^2)，(A^2y？ -ABx？ -BC)/(A 2+B 2)) From the distance formula between two points:

PQ^2=[(B^2x？ Abby? -AC)/(A^2+B^2)-x0]^2

+[(A^2y？ -ABx？ -BC)/(A^2+B^2)-y0]^2

=[(-A^2x？ Abby? -AC)/(A^2+B^2)]^2

+[(-ABx？ -B^2y？ -BC)/(A^2+B^2)]^2

=[A(-By？ -C-Ax？ )/(A^2+B^2)]^2

+[B(-Ax？ -C-By？ )/(A^2+B^2)]^2

=A^2(Ax？ +By？ +C)^2/(A^2+B^2)^2

+B^2(Ax？ +By？ +C)^2/(A^2+B^2)^2

=(A^2+B^2)(Ax？ +By？ +C)^2/(A^2+B^2)^2

=(Ax？ +By？ +C)^2/(A^2+B^2)

So pq = | ax+by+c |/√ (a 2+b 2), and the formula is proved.