Distance from a point to a line 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
Incircle and excircles of a triangle using the side lengths of the triangle relative to the perimeter—i.e., using the barycentric coordinates given above, normalized to sum to unity—as weights. (The weights are positive so the incenter lies inside the triangle as stated above.) If the three vertices are located at formula_14, formula_15, and formula_16, and the sides opposite these vertices have corresponding lengths formula_7, formula_8, and formula_9, then the incenter is at Denoting the incenter of triangle "ABC" as "I", the distances from the incenter to the vertices combined with the lengths of the triangle sides obey the equation Additionally, where "R" and "r" are the