Mx = (5 – 3) ⁄ (6 – 2) = +1/2 y = mx + b now puting values of mx 5 = (1/2)(6) + b y = (1/2)x + 2 line 1 equatuin The slope of the perpendicular (shortest) line from point_P to Line_I = Mpx = -1 ⁄ (Mɪ) = - 2 ——> for Line_PI The equation of Line_PI is : y = mx + b y = - 2x + b ... substitute point_P 6 = - 2(2) + b b = 10 y = - 2x + 10 Line_PI equation The intersection between Line_I and Line_PI is located at : y = y (1/2)x + 2 = - 2x + 10 x = 3.2 and the y_value at that x_location is: y = (1/2)x + 2 Line_I equation y = (1/2)(3.2) + 2 y = 3.6 y = - 2x + 10 Line_PI equation y = - 2(3.2) + 10 y = 3.6 so the intersection is at point_X = (3.2, 3.6) and the distance from point_P to point_X is : d = under root [ (∆x)² + (∆y)² ] d = under root [ (3.2 – 2)² + (6 – 3.6)² ] d = under root 7.2 = under root (36 ⁄ 5) = 6(under root 5) ⁄ 5 = 2.68
