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

Answer 1

Here,

First need to know perpendicular formula.

ie. m₁m₂=-1.----------(1)

Now the given equation is 3x+3y+4=0. ------------(2)

then, y=(-3x-4)/3

=>y=-x-4/3

now comparing with standard equation,

y=mx+b

we get,

slope m₁=-1

now by using equation (1)

m₁m₂=-1,

we get

m₂.(-1) = -1

=> m₂=1

Now using (y₁-y₀)=m₂(x₁-x₀) for (-3,5) and slope 1 according to question and above solution we get,

=> y₁-5 = 1 *(x₁-(-3))

=> y₁-x₁ = 8 -----------(3)

By solving equation 2 and equation 3 we get

x= 34/3

y= 10/3

closest point is (34/3, 10/3)

To verify this solution please check by plugging the value on any equation (2) or (3)..

Answer 2

The distance from the given point to an arbitrary x-y location is given by:

d² = (x + 3)² + (y - 5)²

For the arbitrary location to be on the given line, we need to solve the first equation for y or x and then substitute.

Solve the original equation for y:

3x + 3y + 4 = 0

3y = -3x - 4

y = -x - 4/3

Substitute for y in the distance formula:

d² = (x + 3)² + (-x - 4/3 - 5)²

d² = (x + 3)² + (-x - 19/3)²

Expand and simplify:

d² = x² + 6x + 9 + x² + 38x/3 + 361/9

d² = 2x² + 56x/3 + 442/9

To find the minimum d (which is also the minimum d²), take the first derivative and set = 0

(d²)' = 4x + 56/3 = 0

4x = -56/3

x = - 56/12 = -14/3

from the original equation:

y = -x - 4/3

y = -(-14/3) - 4/3

y = 10/3

The desired point is then (-14/3, 10/3)