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2 years ago

Given 2 points (-2,6) and (5,-8) what is the slope, distance, and midpoint

1 answer:

5 0

(-2,6) (5,-8)

slope = (y2 - y1) / (x2 - x1)
slope = (-8 - 6) / (5 - (-2) = -14/7 = -2 <==

midpoint = (x1 + x2)/2 , (y1 + y2)/2
m = (-2 + 5)/2 , (6 - 8)/2
m = (3/2, -1) <===

distance = sqrt ((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt ((5 - (-2)^2 + (-8 - 6)^2)
d = sqrt ((5 + 2)^2 + (-14^2))
d = sqrt (7^2 + 14^2)
d = sqrt (49 + 196)
d = sqrt 245
d = 15.65 <==

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Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$ .

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$ , then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$ .

Point $F$ would be inside this circle if $d(J,\, F) < r$ . (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$ . (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$ . (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$ :

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$ .

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$ , is smaller than $d(J,\, F)$ . Therefore, point $F$ would be outside this circle.

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