Find the point on the line 3x+5y+1=0 which is closest to the point (5,−3).

Answer 1

Hi,

I can show in a diagram with congruent triangles

that d=distance from point x₀,y₀ to line y=mx+b

always follows formula d=|(mx₀+b)-y₀|/(1+m²)^(1/2)

in words absolute value of (mx₀+b)-y₀

and that value divided by the square root of 1+m²

here, m=-3/5, b=-1/5, x₀=5,y₀=-3

my calculations show d=1/(34)^(1/2)

that is, 1 divided by square root 34

Answer 2

The shortest distance is on the normal (perpendicular) to the line.

3x + 5y +1 = 0

5y = -3x - 1

y = -3x/5 - 1/5 slope of the line is -3/5, slope of normal is 5/3

Using point-slope form

y + 3 = (5/3)(x -5)

Determine point of intersection of above equation and the given equation.

Then use Distance Formula to determine the distance between the two points.