Chapter 4.6, Problem 10PSA

a.

To determine

To find: If given points are collinear.

Answer to Problem 10PSA

Given points are non-collinear.

Explanation of Solution

Given information:

Three points are given as $\left(-6,5\right),\left(1,7\right)\text{ and }\left(15,10\right)$ .

Calculation:

Let us suppose given points are $P=\left(-6,5\right),T=\left(1,7\right)\text{ and }V=\left(15,10\right)$ .

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{PT}=\frac{7-5}{1-\left(-6\right)}=\frac{2}{7}\\ \text{Slope of line }\overleftrightarrow{TV}=\frac{10-7}{15-1}=\frac{3}{14}\end{array}$

  $\text{Since, Slope of line }\overleftrightarrow{PT}\ne \text{Slope of line }\overleftrightarrow{TV}$ .

Therefore, given points are non-collinear.

b.

To determine

To find: If given points are collinear.

Answer to Problem 10PSA

Given points are collinear.

Explanation of Solution

Given information:

Three points are given as $\left(74,20\right),\left(50,16\right)\text{ and }\left(2,8\right)$ .

Calculation:

Let us suppose given points are $P=\left(74,20\right),T=\left(50,16\right)\text{ and }V=\left(2,8\right)$ .

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{PT}=\frac{16-20}{50-74}=\frac{1}{6}\\ \text{Slope of line }\overleftrightarrow{TV}=\frac{8-16}{2-50}=\frac{1}{6}\end{array}$

  $\text{Since, Slope of line }\overleftrightarrow{PT}=\text{Slope of line }\overleftrightarrow{TV}$ .

Therefore, given points are collinear.