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
