Distance from a point to a line
Introduction In Euclidean geometry, the '''distance from a point to a line''' is the shortest distance from a given point to any point on an infinite straight line. It is the perpendicular distance of the point to the line, the length of the line segment which joins the point to nearest point on the line. The algebraic expression for calculating it can be derived and expressed in several ways.Knowing the distance from a point to a line can be useful in various situations—for example, finding the shortest distance to reach a road, quantifying the scatter on a graph, etc. In Deming regression, a type of linear curve fitting, if the dependent and independent variables have equal variance this results in orthogonal regression in which the degree of imperfection of the fit is measured for each data point as the perpendicular distance of the point from the regression line.
Line defined by an equation In the case of a line in the plane given by the equation , where ,  and  are real constants with  and  not both zero, the distance from the line to a point  is:The point on this line which is closest to  has coordinates::'''Horizontal and vertical lines'''In the general equation of a line, ,  and  cannot both be zero unless  is also zero, in which case the equation does not define a line. If  and , the line is horizontal and has equation . The distance from  to this line is measured along a vertical line segment of length  in accordance with the formula. Similarly, for vertical lines (''b'' = 0) the distance between the same point and the line is , as measured along a horizontal line segment.
Line defined by two points If the line passes through two points  and  then the distance of  from the line is::The denominator of this expression is the distance between  and . The numerator is twice the area of the triangle with its vertices at the three points, ,  and . See: . The expression is equivalent to , which can be obtained by rearranging the standard formula for the area of a triangle: , where  is the length of a side, and  is the perpendicular height from the opposite vertex.
Line defined by point and angle If the line passes through the point  with angle , then the distance of some point  to the line is:
Proofs ===An algebraic proof===This proof is valid only if the line is neither vertical nor horizontal, that is, we assume that neither  nor  in the equation of the line is zero.The line with equation  has slope , so any line perpendicular to it will have slope  (the negative reciprocal). Let  be the point of intersection of the line  and the line perpendicular to it which passes through the point (, ). The line through these two points is perpendicular to the original line, so  :Thus,and by squaring this equation we obtain::Now consider,:using the above squared equation. But we also have,:since  is on .Thus,:and we obtain the length of the line segment determined by these two points,:===A geometric proof===Diagram for geometric proofThis proof is valid only if the line is not horizontal or vertical.Drop a perpendicular from the point ''P'' with coordinates (''x''0, ''y''0) to the line with equation ''Ax'' + ''By'' + ''C'' = 0.Label the foot of the perpendicular ''R''. Draw the vertical line through ''P'' and label its intersection with the given line ''S''. At any point ''T'' on the line, draw a right triangle ''TVU'' whose sides are horizontal and vertical line segments with hypotenuse ''TU'' on the given line and horizontal side of length |''B''| (see diagram). The vertical side of ∆''TVU'' will have length |''A''| since the line has slope -''A''/''B''.∆''PRS'' and ∆''TVU'' are similar triangles, since they are both right triangles and ∠''PSR'' ≅ ∠''TUV'' since they are corresponding angles of a transversal to the parallel lines ''PS'' and ''UV'' (both are vertical lines). Corresponding sides of these triangles are in the same ratio, so::If point ''S'' has coordinates (''x''0,''m'') then |''PS''| = |''y''0 - ''m''| and the distance from ''P'' to the line is::Since ''S'' is on the line, we can find the value of m,:and finally obtain::A variation of this proof is to place V at P and compute the area of the triangle ∆''UVT'' two ways to obtain that where D is the altitude of ∆''UVT'' drawn to the hypotenuse of ∆''UVT'' from ''P''. The distance formula can then used to express , , and in terms of the coordinates of P and the coefficients of the equation of the line to get the indicated formula.===A vector projection proof===Diagram for vector projection proofLet ''P'' be the point with coordinates (''x''0, ''y''0) and let the given line have equation ''ax'' + ''by'' + ''c'' = 0.Also, let ''Q'' = (''x''1, ''y''1) be any point on this line and '''n''' the vector (''a'', ''b'') starting at point ''Q''. The vector '''n''' is perpendicular to the line, and the distance ''d'' from point ''P'' to the line is equal to the length of the orthogonal projection of  on '''n'''. The length of this projection is given by::Now,: so  and thus:Since ''Q'' is a point on the line, , and so,:Although the distance is given as a modulus, the sign can be useful to determine which side of the line the point is on, in a sense determined by the direction of normal vector (a,b)
Another formula It is possible to produce another expression to find the shortest distance of a point to a line. This derivation also requires that the line be not vertical or horizontal.The point P is given with coordinates ().The equation of a line is given by . The equation of the normal of that line which passes through the point P is given .The point at which these two lines intersect is the closest point on the original line to the point P. Hence::We can solve this equation for ''x'',:The ''y'' coordinate of the point of intersection can be found by substituting this value of ''x'' into the equation of the original line,:Using the equation for finding the distance between 2 points, , we can deduce that the formula to find the shortest distance between a line and a point is the following::Recalling that ''m'' = -''a''/''b'' and ''k'' = - ''c''/''b'' for the line with equation ''ax'' + ''by'' + c = 0, a little algebraic simplification reduces this to the standard expression.
Vector formulation Illustration of the vector formulation.The equation of a line can be given in vector form:: Here  is a point on the line, and  is a  unit vector in the direction of the line. Then as scalar ''t'' varies,  gives the locus of the line.The distance of an arbitrary point  to this line is given by: This formula can be derived as follows:  is a vector from  to the point . Then  is the projected length onto the line and so:is a vector that is the projection of  onto the line and represents the point on the line closest to . Thus:is the component of  perpendicular to the line. The distance from the point to the line is then just the norm of that vector. This more general formula is not restricted to two dimensions.
Another vector formulation If the vector space is orthonormal and if the line goes through point  and has a direction vector , the distance between point  and the line is: Note that cross products only exist in dimensions 3 and 7.
See also * Hesse normal form* Line-line intersection* Distance between two lines* Distance from a point to a plane* Skew lines#Distance
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