Please help me with this question

Question

Find the point on the line 5𝑥+2𝑦−1=0 which is closest to the point (4,−4)

Answer 1

One way to do the problem (without calculus): Find the equation of the line passing through (4,-4) that is perpendicular to given line. Note the slope of the given line is -5/2 so the equation of the perpendicular is 2/5. Now find the point of intersection of the original line and the perpendicular passing through (4,-4). That is the point on the original line that is closest to (4,-4).

See:

desmos.com/calculator/vsk5xdqho6

Answer 2

The equation of the line is 5x + 2y - 1 = 0 or 5x + 2y = 1

You can move the point to (0,0) and the line by plugging in

(4, -4) into the equation of the line 5x + 2y and getting 5(4) + 2(-4) = 12

We now seek the distance where (0, 0) is closest to 5x + 2y = 1 - 12 = -11

This is 11/sqrt(5^2+2^2) = 11/sqrt(29) = 11 sqrt(29)/29

The Calculus way:

5x + 2y - 1 = 0

y = 1/2 - 5/2x

We seek to minimize the distance from (x, y) to (4, -4) or (x, 1/2-5/2x) to (4, -4)

The distance is sqrt((x-4)^2 + (y- -4)^2) = sqrt((x-4)^2 + (1/2-5/2x - -4)^2) =

sqrt((x-4)^2 + (9/2-5/2x)^2)

It is easier to minimize the distance squared equals (x-4)^2 + (9/2+5/2x)^2

We set the derivative to 0

2(x-4) + 2(-5/2)(9/2-5/2x) = 0

2x - 8 - 45/2 + 25/2x = 0

29/2x = 61/2

x = 61/29

y = 1/2 - 5/2x

y = 1/2 - 5/2(61/29) = -4 22/29

The distance is sqrt((61/29-4)^2 + (-4 22/29 - -4)^2) = sqrt((-55/29)^2 + (-22/29)^2) =

11/29 sqrt((-5)^2 + (-2)^2) = 11/29 * sqrt(29) = 11 sqrt(29)/29 or 11/sqrt(29)

Answer 3

The formula for the distance from a point to a line is this:

d=(ax₀ + by₀ + c)/(sqrt(a² + b²)

where the given line is ax + by + c = 0 and (x₀,y₀) is the given point.

This is the perpendicular distance from the point to the line which is the shortest distance.

You plug in the numbers.

P.S.

The derivation of this formula is much too long to include here...sorry.