Question

Find the distance from P to l. . Line l contains points (6,5) and (2,3). . Point P has coordinates (2,6).

Answer 1

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