Find the point on the line 4x + y = 9 that is closest to the point (-4,2)

Answer 1

We have the line y = 9 - 4*x and the point (-4, 2). The distance from the point (-4, 2) to an arbitrary point (x, y) is:

d = sqrt[ (-4 - x)^2 + (2 - y)^2 ].

A point on the line will have y = 9 - 4*x so we can substitute for y:

d = sqrt[ (-4 - x)^2 + (2 - 9 + 4*x)^2 ] = sqrt[ (16 + 8x + x^2) + (49 - 56x + 16*x^2) ]

= sqrt[ 17*x^2 - 48*x + 65 ]

Taking the derivative we get:

d(d)/dx = (1/2)*(34*x - 48)

To find the minimum set the derivative to zero

34*x - 48 = 0 => x = 48/34 = 1.412

The corresponding y value is y = 9 - 4*x = 3.353 so the closest point on the line is:

(1.412, 3.353)

Answer 2

Calculus is not really required to do this problem. The slope of a line perpendicular to 4x+y=9 will be 1/4. The equation of such a line passing through (-4,2) is y-2 = 1/4 (x+4). Solve the system of equations for x to find point of intersection, then find distance between that point and (-4,2). This works because shortest distance between two points is along the perpendicular.

You could use calculus to do the same by realizing that the coordinates of any point on the given line can be represented as (x, -4x+9). Then the distance can be found by d = [ (x +4)² + (-4x + 9 -2)²]^1/2..

Find the derivative of d with respect to x, set it equal to zero to locate critical number(s). Show that that critical number generates a min, substitute into d to find the shortest distance.

You actually to not have to include the radical as part of the distance formula, because it will achieve min/max where the radicand achieves min/max.

If you do include the radical, derivative is given by:

2(x+4) + 2(-4x+7)(-4)

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2 √etc.

That will only equal zero when the numerator equals zero. So simplify the numerator, set equal to zero, solve for x, etc.