2.4: Applied Examples

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Page ID: 45179

Victoria Dominguez, Cristian Martinez, & Sanaa Saykali
Citrus College via ASCCC Open Educational Resources Initiative

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In this section, apply the distance formula \(d = \sqrt{(x_2 − x_1) ^2 + (y_2 − y_1) ^2}\) to find the lengths of line segments.

Note: Three points \(A\), \(B\), and \(C\) are collinear, or in other words, the three points lie on the same line, if the sum of the lengths of any two line segments connecting the points, is equal to the length of the remaining line segment. That is, \(AB + BC = AC\) or, \(AB + BC = AC\) or, \(AB + AC = BC\) or \(AC + BC = AB\).

Example 2.4.1

Determine if the given three points are collinear.

\(A(10, −4)\quad B(8, −2) \quad C(2, 4)\)

Solution

First find segments \(AB\), \(BC\), and \(AC\). To do so, find the distance between the points \(A\) and \(B\), \(B\) and \(C\), \(A\) and \(C\).

\(\begin{aligned} \text{Segment AB }&=\text{ The distance between point A and Point B } \\ &= \sqrt{(8 − 10)^2 + [−2 − (−4)]^2} \\ &= \sqrt{(−2)^2 + (2)^2} \\&= \sqrt{ 8}\\&= 2\sqrt{2} \end{aligned}\)

\(\begin{aligned} \text{Segment BC }&=\text{ The distance between point B and Point C } \\ &= \sqrt{(2 − 8)^2 + [4 − (−2)]^2 }\\ &= \sqrt{(−6)^2 + (6)^2} \\&= \sqrt{ 72 }\\&= 6\sqrt{ 2}\end{aligned}\)

\(\begin{aligned} \text{Segment AC }&=\text{ The distance between point A and Point C }\\&= \sqrt{(2 − 10)^2 + [4 − (−4)]^2} \\&= \sqrt{(−8)^2 + (8)^2 }\\&= \sqrt{ 128 }\\&= 8\sqrt{ 2}\end{aligned}\)

Thus,

\(\begin{aligned} AB + BC &= 2\sqrt{ 2} + 6\sqrt{ 2 }\\&= 8\sqrt{ 2 } \\&= AC \end{aligned}\)

Since Thus, \(AB + BC = AC\) then three points are collinear.

Exercise 2.4.1

Determine if the following points are collinear.
1. \(A(4,-1)\quad B(5,-2) \quad C(1,2)\)
2. \(A(2,-2)\quad B(3,1)\quad C(2,1)\)