Chapter 1.1, Problem 52E

i.

To determine

To calculate: Three or more points are collinear when they all lie on the same line. Use the steps below to determine whether the set of points $\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and the set of points $\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$ are collinear. For each set of points, use the Distance Formula to find the distance from $A$ to $B$ , $B$ to $C$ , $A$ to $C$ . What relationship exists among these distances for each set of points?

Answer to Problem 52E

Distances are 3,5,4 and $\sqrt{10},\sqrt{10},2\sqrt{10}$

Explanation of Solution

Given information:

Given set of points are $\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and $\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$

Let $L_1=\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and $L_2=\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$

Concept used:

Distance between two points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is given by $d_{AB}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Using above formula distance of $L_1$ from $A$ to $B$ is

  $\begin{array}{l}{d}_{AB}=\sqrt{{\left(2-2\right)}^2+{\left(3-6\right)}^2}\\ d_{AB}=\sqrt{9}\\ d_{AB}=3\end{array}$

Distance of $L_1$ from $B$ to $C$ is

  $\begin{array}{l}{d}_{BC}=\sqrt{{\left(2-6\right)}^2+{\left(6-3\right)}^2}\\ d_{BC}=\sqrt{16+9}\\ d_{BC}=\sqrt{25}\\ d_{BC}=5\end{array}$

Distance of $L_1$ from $A$ to $C$ is

  $\begin{array}{l}{d}_{AC}=\sqrt{{\left(2-6\right)}^2+{\left(3-3\right)}^2}\\ d_{AC}=\sqrt{16}\\ d_{AC}=4\end{array}$

Using above formula distance of $L_2$ from $A$ to $B$ is

  $\begin{array}{l}{d}_{AB}=\sqrt{{\left(8-5\right)}^2+{\left(3-2\right)}^2}\\ d_{AB}=\sqrt{10}\end{array}$

Distance of $L_2$ from $B$ to $C$ is

  $\begin{array}{l}{d}_{BC}=\sqrt{{\left(5-2\right)}^2+{\left(2-1\right)}^2}\\ d_{AB}=\sqrt{10}\end{array}$

Distance of $L_2$ from $A$ to $C$ is

  $\begin{array}{l}{d}_{AC}=\sqrt{{\left(8-2\right)}^2+{\left(3-1\right)}^2}\\ d_{AC}=\sqrt{36+4}\\ d_{AC}=\sqrt{40}\\ d_{AC}=2\sqrt{10}\end{array}$

Conclusion:

The set of points formed by $L_1$ satisfy the Pythagorean Theorem.

The set of points formed by $L_2$ have distances that sum up to each other.

ii.

To determine

To calculate: Three or more points are collinear when they all lie on the same line. Use the steps below to determine whether the set of points $\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and the set of points $\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$ are collinear. Plot each set of points in the Cartesian plane. Do all the points of either set appear to lie on the same line?

Answer to Problem 52E

For $L_1$ all point do not lie on same line.

For $L_2$ all point lie on a straight line.

Explanation of Solution

Given information:

Given set of points are $\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and $\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$

Let $L_1=\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and $L_2=\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$

To plot the point $(x,y)$ , imagine a vertical line through $x$ on $x-axis$ and a horizontal line through $y$ on the $y-\text{axis}$ . Intersection of these two lines is the point $(x,y)$

Graph of $L_1$ with points $L_1=\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$

  

Graph of $L_2$ with points $L_2=\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$

  

Conclusion: For $L_1$ all point do not lie on same line.

For $L_2$ all point lie on a straight line.

iii.

To determine

To calculate: Three or more points are collinear when they all lie on the same line. Use the steps below to determine whether the set of points $\left\{A\left(2,3\right),B\left(2,6\right),C\left(6,3\right)\right\}$ and the set of points $\left\{A\left(8,3\right),B\left(5,2\right),C\left(2,1\right)\right\}$ are collinear. Compare your conclusion from part (a) with the conclusion you made from the graphs in part (b). Make general statement about how to use the Distance Formula to determine co linearity.

Answer to Problem 52E

Three points $A,B,C$ are collinear when the distance from $A$ to $B$ , $B$ to $C$ sum up the distance from $A$ to $C$ .

Explanation of Solution

from part (a) and part(b) we concluded that for 3 points, if sum of distances between any two points is equal to the distance between other points. so three points $A,B,C$ are collinear when the distance from $A$ to $B$ , $B$ to $C$ sum up the distance from $A$ to $C$ .