Respuesta :

The distance from a point to a line is
distance(ax+by+c=0,(x0,y0)) = abs(ax0+by0+c)/sqrt(a^2+b^2)
So first convert the formula for the line to standard form
5x+2y=4
5x+2y-4=0

a=5, b=2, c=-4, x0=15, y0=-21

abs(5*15+2*(-21)-4)/sqrt(5*5+2*2)
= abs(75-42-4)/sqrt(25+4)
= 29/sqrt(29)
= sqrt(29)
= 5.385165 (approximately)

facundo3141592

We want to find the distance between a given point and a line.

The distance from point A(15, -21) to the line 5x +2y = 4 is 22.47

Remember that the distance between two points (x₁, y₁) and (x₂, y₂) is given by:

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

Here we want to find the distance between the point A(15, -21) and the line 5x + 2y = 4

We can rewrite our line as:

2y = 4 - 5*x

y = (4 - 5*x)/2 = 2 - 2.5*x

Then any point on the line can be written as:

(x, y) = (x, 2 - 2.5*x)

So we want to find the minimum distance between the points A(15, -21)  and (x, 2 - 2.5*x).

We will get:

[tex]d(x) = \sqrt{(15 - x)^2 + (-21 - 2 - 2.5*x)^2} \\\\d(x) = \sqrt{x^2 - 30*x + 225 + 529 + 115*x + 6.25*x^2 }\\\\d(x) = \sqrt{7.25*x^2 +85*x + 754}[/tex]

Here we want to minimize d(x). Notice that if d(x) > 1, then minimizing d(x) is equivalent to minimizing d(x)^2

Then we want to minimize:

[tex]d(x)^2 = 7.25*x^2 +85*x + 754[/tex]

This is just a quadratic equation with a positive leading coefficient, the minimum is at the vertex.

The x-value of the vertex is given by:

x = -(85)/(2*7.25) = -5.86

Then the distance between the point and the line is:

[tex]d(-5.86) = \sqrt{7.25*(-5.86)^2 +85*(-5.86) + 754} = 22.47[/tex]

The distance from point A(15, -21) to the line 5x +2y = 4 is 22.47

If you want to learn more, you can read:

https://brainly.com/question/12082741

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