Verified answer

The closest distance of a line to a point is their perpendicular distance.

SOLUTION USING CALCULUS:

D² = (y₂–y₁)²+(x₂–x₁)²

D² = (y–3)²+(x–5)²

5x+6y–1=0

6y = 1–5x

y = (1/6)–(5/6)x ................ [Plug in this equation to D.]

D² = (y–3)²+(x–5)²

D² = [(1/6)-(5/6)x–3]² + (x–5)²

D² = [-(5/6)x–(17/6)]² + (x–5)²

D² = (25/36)x² + (85/18)x + (289/36) + x² – 10x + 25

D² = (61/36)x² – (95/18)x + (1189/36)

D = √[(61/36)x² – (95/18)x + (1189/36)]

Derivative:

D' = ½ [(61/36)x²–(95/18)x+(1189/36)]^(-½) [(61/18)x – (95/18)]

Equate to zero:

0 = (61/18)x – (95/18)

(61/18)x = 95/18

x = 95/61

So at x = 95/61:

5x + 6y – 1 = 0

5(95/61) + 6y – 1 = 0

y = –69/61

So the point closest to (5,3) is (95/61, –69/61)

Or simply (1.56, –1.13) ......................................... [ Ans.]

SOLUTION USING ANALYTIC GEOMETRY:

y = -(5/6)x + (1/6)

The perpendicular slope is (6/5).

The equation of the perpendicular line that passes through (5,3) is:

y–3=(6/5) (x–5)

y–3 = (6/5)x – 6

y = (6/5)x – 3

Their point of intersection is the point closest. So

y = y

-(5/6)x + (1/6) = (6/5)x – 3

-(5/6)x–(6/5)x = –(19/6)

(61/30)x = 19/6

x = 95/61

Plug in x=95/61 to either equation you will get

y=–69/61

So the point is (95/61, –69/61)

Or simply (1.56, –1.13) ......................................... [ Ans.]