Find the distance from p to l. line l contains points (6,5) and (2,3). point p has coordinates (2,6).

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