GIVEN: that there is a line L that is defined by two points, P1 and P2, and a point Q not on L.

FIND:

1. the point Q0 that is the closest point to Q on the line L.
2. the parametric equation of the second line L2 that contains points Q and Q0.
3. the distance between the point Q and the line L.

SOLUTION:

The parametric equation of L is:

P = P1 + k * (P2 - P1)

where k is a scalar parameter that is equal to 0 when P = P1 and 1 when b>P = P2. The parameter value k can have positive or negative values.

The parametric equation of the second line L2 is:

R = Q + k2 * (Q0 - Q)

where k2 is the scalar parameter of L2 that is equal to 0 when R = Q and 1 when R = Q0. The parameter value k2 can also have positive and negative values.

Now you are ready to solve equations. The first equation is driven by the condition of normalcy. Since L2 ⊥ L, then

(Q0 - Q) • (P2 - P1) = 0

where • is the dot product. The second condition is that Q0 is a point on L:

(Q0 = P1 + k * (P2 - P1)

Solving for k:

k* = ((Q - P1) * (P2 - P1))/ |P2 - P1|2 ...or...
k* = ((Qx-P1x)(P2x-P1x)+(Qy-P1y)(P2y-P1y)) / ((P2x-P1x)2 +(P2y-P1y)2)

Now k* is that value of the line parameter k on line L which yields the coordinates of the point of intersection Q0, the point that lies on both lines L and L2. So we have our answer:

SOLUTION:

The point of intersection (which is also the point on the line L closest to point Q) is given by:

k* = ((Q - P1) * (P2 - P1))/ |P2 - P1| ...or...
k* = ((Qx-P1x)(P2x-P1x)+(Qy-P1y)(P2y-P1y)) / ((P2x-P1x)2 +(P2y-P1y)2)

and therefore the point of intersection is

Q = P1 + k* * (P2 - P1)

The parametric equation of the normal line L2 is:

R = Q + k2 * (Q0 - Q)

Finally, the distance between the point and the line is equal to

|Q0 - Q|
= SQRT((Q0x-Qx)2 +(Q0y-Qy)2)

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