Chapter B, Problem 6E

(a)

To determine

To calculate: To show that the points $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$ are collinear by showing that $\left|AB\right|+\left|BC\right|=\left|AC\right|$

Answer to Problem 6E

It isproved that points are collinear by showing that $\left|AB\right|+\left|BC\right|=\left|AC\right|$

Explanation of Solution

Given information: Points are $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$

Formula Used:

Distance between two points is given as

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

Calculation:

Points are given as

  $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$

Length of $AB$ is calculated as follows:

  $\begin{array}{l}\left|AB\right|=\sqrt{{\left(3-\left(-1\right)\right)}^2+{\left(11-3\right)}^2}\\ \left|AB\right|=\sqrt{4^2+8^2}\\ \left|AB\right|=\sqrt{16+64}\\ \left|AB\right|=\sqrt{80}\\ \left|AB\right|=4\sqrt{5}\end{array}$

Length of $BC$ is calculated as follows:

  $\begin{array}{l}\left|BC\right|=\sqrt{{\left(5-3\right)}^2+{\left(15-11\right)}^2}\\ \left|BC\right|=\sqrt{2^2+4^2}\\ \left|BC\right|=\sqrt{4+16}\\ \left|BC\right|=\sqrt{20}\\ \left|BC\right|=2\sqrt{5}\end{array}$

Length of $AC$ is calculated as follows:

  $\begin{array}{l}\left|AC\right|=\sqrt{{\left(5-\left(-1\right)\right)}^2+{\left(15-3\right)}^2}\\ \left|AC\right|=\sqrt{6^2+{12}^2}\\ \left|AC\right|=\sqrt{36+144}\\ \left|AC\right|=\sqrt{180}\\ \left|AC\right|=6\sqrt{5}\end{array}$

Now, calculating $\left|AB\right|+\left|BC\right|$ ,

  $\begin{array}{l}\left|AB\right|+\left|BC\right|=4\sqrt{5}+2\sqrt{5}\\ \left|AB\right|+\left|BC\right|=6\sqrt{5}\\ \left|AB\right|+\left|BC\right|=\left|AC\right|\end{array}$

Thus, points A, B and C are collinear

Conclusion:

Hence, its proved that points are collinear by showing that $\left|AB\right|+\left|BC\right|=\left|AC\right|$

(b)

To determine

To calculate: To show that the points $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$ are collinear by using slopes

Answer to Problem 6E

It isproved that points are collinear by showing that slopes are equal

Explanation of Solution

Given information: Points are $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$

Formula Used:

Slope of line joining two points is calculated as

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Points are given as

  $A\left(-1,3\right)$

  $B\left(3,11\right)$ and $C\left(5,15\right)$

Slope of line $AB$ is calculated as follows:

  $\begin{array}{l}{m}_{AB}=\frac{11-3}{3-\left(-1\right)}\\ m_{AB}=\frac{8}{4}\\ m_{AB}=2\end{array}$

Slope of line $BC$ is calculated as follows:

  $\begin{array}{l}{m}_{BC}=\frac{15-11}{5-3}\\ m_{BC}=\frac{4}{2}\\ m_{BC}=2\end{array}$

Slope of line $AC$ is calculated as follows:

  $\begin{array}{l}{m}_{AC}=\frac{15-3}{5-\left(-1\right)}\\ m_{AC}=\frac{12}{6}\\ m_{AC}=2\end{array}$

Since $m_{AB}=m_{BC}=m_{AC}=2$

Thus, points A, B and C are collinear

Conclusion:

Hence, its proved that points are collinear by showing that slope of lines are equal