Find the point on the line −2x+3y−2=0 which is closest to the point (1,−5) .

Question

Please help

Answer 1

Two basic ways to do the problem.

No Calculus:

y = (2/3)x+2/3

Slope of line perpendicular to the given line is: -3/2

Passing through (1, -5) has equation:

y + 5 = (-3/2)(x - 1)

y = (-3/2)x - 7/2

The point of intersection of those two lines will be the closest point since it lies on the perpendicular..

(2/3)x + 2/3 = (-3/2)x - 7/2

4x + 4 = -9x -21

13x = -25

x = -25/13

y = -8/13

Using calculus:

Any point on the given line has coordinates (x, (2/3)x + 2/3))

The distance from that general point to (1, -5) is given by the function:

d² = ( x -1)² + ( (2/3)x + 2/3 +5)²

Taking the derivative of d with respect to x:

2dd' = 2(x-1) + 2((2/3)x+17/3)(2/3)

d' = [(x-1) + (4/9)x+34/9]/d

That is equal to zero when the numerator equals zero:

x - 1 +(4/9)x + 34/9 = 0

9x - 9 +4x + 34 = 0

13x = -25

x = -25/13

y = -8/13

In this case 1st or 2nd derivative test can be used to show that x = -25/13 given a minimum distance.

If you just needed to get the distance from the point to the line, there is a formula for that:

See:

desmos.com/calculator/wcmpxrbzxp

Answer 2

The equation of the line is -2x + 3y - 2 = 0.

Let's solve the above equation for y, yielding y = (2/3)x + 2/3.

So it's a line of slope 2/3 (going from lower left to upper right) with a y-intercept of 2/3.

Sketch this line, and sketch the point (1, -5), and you'll see that the point on the line closest to (1, -5) is approximately (-2, -1), down in the "third quadrant".

Now let's get a more exact answer.

Given any x value, the point on the line with that x value is the point (x, (2/3)x + 2/3).

The distance from this point to the point (1, -5) is given by the Pythagorean theorem.

[Quick review of how to use this theorem to get the distance between two points, e.g., (x1, y1) and (x2, y2).

The distance is ((x1 - x2)² + (y1 - y2)²)^.5 in other words, sqrt((x1 - x2)² + (y1 - y2)²).]

So the distance between point (x, (2/3)x + 2/3) and point (1, -5) is

((x - 1)² + ((2/3)x + 2/3 -(-5))²)^.5 which we can simplify to ((x - 1)² + ((2/3)x + 17/3)²)^.5

So the distance between the point on the line with a given x coördinate (i.e., the point (x, (2/3)x + 2/3)) and the point (1, -5) is given by the function f(x) = ((x - 1)² + ((2/3)x + 17/3)²)^.5

We want to find the x value that causes this function to take on its minimum value.

f is a continuous function, and a continuous function takes on its minimum and maximum values, if any, at the points where the function is horizontal, i.e., at the points where the function's derivative is zero. (The top of a hill, and the bottom of a valley, are always flat.)

Since f(x) = ((x - 1)² + ((2/3)x + 17/3)²)^.5

we have f'(x) = .5((x - 1)² + ((2/3)x + 17/3)²)^-.5(2(x-1) + 2((2/3)x + 17/3)(2/3))

which we can simplify to

f'(x) = .5((x - 1)² + ((2/3)x + 17/3)²)^-.5(2x - 2 + (8/9)x + 68/9)

which we can simplify further to

f'(x) = .5((x - 1)² + ((2/3)x + 17/3)²)^-.5((26/9)x + 50/9)

Visualize this f'(x) as a gigantic fraction with .5((26/9)x + 50/9) up in the numerator and

((x - 1)² + ((2/3)x + 17/3)²)^.5 down in the denominator. I'm sorry I can't type it here that way.

To make this f'(x) = 0, all we have to do is make the factor (in the numerator) ((26/9)x + 50/9) = 0.

The solution that makes it zero is x=-25/13.

So f'(-25/13) = 0, and x=-25/13 is a minimum point for the function f, since f has no maximum points. f just gets smoothly bigger and bigger all the way to infinity without ever reaching a local maximum.

So the point on our original line -2x + 3y -2 = 0 that is closest to the point (1, -5) is the point whose x coördinate is -25/13 (i.e., the point (-25/13, -8/13), which is tolerably close to our original estimate of (-2, -1)).

So our answer is (-25/13, -8/13). Good night.