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Chapter 51: Problem 890

Show that the point \(\mathrm{P}_{1}(2,2,3)\) is equidistant from the points \(\mathrm{P}_{2}(1,4,-2)\) and \(\mathrm{P}_{3}(3,7,5)\)

Short Answer

Expert verified
To show that point P1(2,2,3) is equidistant from points P2(1,4,-2) and P3(3,7,5), we calculate the distance between P1 and P2 and P1 and P3 using the distance formula for 3D space. We find that \(d_{P1P2} = \sqrt{30}\) and \(d_{P1P3} = \sqrt{30}\). Since the distances are equal, point P1 is indeed equidistant from both points P2 and P3.

Step by step solution

01

Calculate the distance between P1 and P2

Using the distance formula: \[d_{P1P2} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\] where P1(2,2,3) and P2(1,4,-2), plug in the coordinates: \[d_{P1P2} = \sqrt{(1 - 2)^2 + (4 - 2)^2 + (-2 - 3)^2}\]
02

Simplify the equation for d_{P1P2}

Simplify the equation for d_{P1P2}: \[d_{P1P2} = \sqrt{(-1)^2 + (2)^2 + (-5)^2}\] \[d_{P1P2} = \sqrt{1 + 4 + 25}\] \[d_{P1P2} = \sqrt{30}\]
03

Calculate the distance between P1 and P3

Using the distance formula: \[d_{P1P3} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\] where P1(2,2,3) and P3(3,7,5), plug in the coordinates: \[d_{P1P3} = \sqrt{(3 - 2)^2 + (7 - 2)^2 + (5 - 3)^2}\]
04

Simplify the equation for d_{P1P3}

Simplify the equation for d_{P1P3}: \[d_{P1P3} = \sqrt{(1)^2 + (5)^2 + (2)^2}\] \[d_{P1P3} = \sqrt{1 + 25 + 4}\] \[d_{P1P3} = \sqrt{30}\]
05

Compare the distances

Now that we have both distances, d_{P1P2} and d_{P1P3}, we compare them: \[ d_{P1P2} = d_{P1P3}\] \[\sqrt{30} = \sqrt{30}\] Since the distances between point P1 and points P2 and P3 are equal, we can conclude that point P1 is equidistant from both points P2 and P3.

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