Chapter 9.5, Problem 19PSB

a

To determine

To show:ThePoints are collinear using the distance formula.

Explanation of Solution

Given information:Here, we have given that points are $(1,2)$ , $(4,6)$ and $(10,14)$

Formula used: Distance formula: d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Proof:Let the points be A $(1,2)$ , B $(4,6)$ and C $(10,14)$ .

Points are collinear when $AB+BC=AC$ .

  $\begin{array}{l}AB=\sqrt{{(4-1)}^2+{(6-2)}^2}\\ AB=\sqrt{{(3)}^2+{(4)}^2}\\ AB=\sqrt{9+16}\\ AB=\sqrt{25}\\ AB=5\end{array}$

Now,

  $\begin{array}{l}BC=\sqrt{{(10-4)}^2+{(14-6)}^2}\\ BC=\sqrt{{(6)}^2+{(8)}^2}\\ BC=\sqrt{36+64}\\ BC=\sqrt{100}\\ BC=10\end{array}$

And,

  $\begin{array}{l}AC=\sqrt{{(10-1)}^2+{(14-2)}^2}\\ AC=\sqrt{{(9)}^2+{(12)}^2}\\ AC=\sqrt{81+144}\\ AC=\sqrt{225}\\ AC=15\end{array}$

As, $AB+BC=AC$

  $\Rightarrow$ Points are collinear.

b

To determine

To show: Points are collinear using slope.

Explanation of Solution

Given information:

Here, we have given that points are $(1,2)$ , $(4,6)$ and $(10,14)$

Formula used: Slope of a line = $\frac{y_2-y_1}{x_2-x_1}$ .

Proof:Let the points be A $(1,2)$ , B $(4,6)$ and C $(10,14)$ .

Points are collinear when, slope of AB = slope of BC.

Slope of AB = $\frac{6-2}{4-1}$

  $\Rightarrow \frac{4}{3}$

Now, Slope of BC = $\frac{14-6}{10-4}$

  $\begin{array}{l}\Rightarrow \frac{8}{6}\\ \Rightarrow \frac{4}{3}\end{array}$

Now, as Slope pf AB = Slope pf BC

  $\Rightarrow$ Points are collinear.