Find the point on the line 4𝑥+5𝑦+3=0 which is closest to the point (3,3).

Answer 1

For a point (x₀,y₀), the shortest distance to the line ax+by+c=0 is to the point (x₁,y₁):

x₁ = [b(bx₀-ay₀)-ac]/(a²+b²)

y₁ = [a(-bx₀+ay₀)-bc]/(a²+b²)

https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

In this problem (x₀,y₀)=(3,3) and a=4, b=5, and c=3

x₁ = [5(5•3-4•3)-4•3]/(4²+5²)=[5(15-12)-12]/(16+25)

=[5(3)-12]/41=3/41

y₁ = [4(-5•3+4•3)-5•3]/(4²+5²)=[4(-15+12)-15]/(16+25)

=[4(-3)-15]/41=-27/41

Solution: (3/41,-27/41)



Answer 2

Find the equation of the line which is perpendicular to the given line and passes through (3,3)

The intersection of this perpendicular and the given line will be the point you need.

If you are unable to follow those directions to the answer, comment so that I can come back and give more help.