Question

Given (9,4) and (x,−11), find all x such that the distance between these two points is 17. Separate multiple answers with a comma.

Answer 1

Answer:

The points are (1, -11) and (17, -11).

Step-by-step explanation:

The distance between points (x1, y1) and (x2, y2) is

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

We have values for x1, y1, and y2. We leave x for x2, and we solve for x.

$17 = \sqrt{(x - 9)^2 + (-11 - 4)^2}$

Square both sides.

$289 = (x - 9)^2 + (-11 - 4)^2$

$289 = x^2 - 18x + 81 + 225$

$x^2 - 18x + 17 = 0$

$(x - 1)(x - 17) = 0 [/tex}

[tex] x - 1 = 0$   or   $x - 17 = 0$

x = 1 or x = 17

The points are (1, -11) and (17, -11).

Answer 2

Using distance between two points, it is found that the values are:  x = 17, x = 1.

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Distance between two points:

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$. The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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• Points (9,4) and (x, -11).
• Distance of 17, thus $D = 17$

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sqrt{(x - 9)^2 + (-11 - 4)^2} = 17$

$(\sqrt{(x - 9)^2 + (-11 - 4)^2})^2 = 17^2$

$(x - 9)^2 + (-15)^2 = 289$

$x^2 - 18x + 81 + 225 = 289$

$x^2 - 18x + 17 = 0$

Quadratic equation with $a = 1, b = -18, c = 17$.

Applying Bhaskara:

$\Delta = b^2 - 4ac = (-18)^2 - 4(1)(17) = 256$

$x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-(-18) + \sqrt{256}}{2} = 17$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-(-18) - \sqrt{256}}{2} = 1$

The values are: x = 17, x = 1.

A similar problem is given at https://brainly.com/question/22532602