Chapter 13.2, Problem 34WE

To determine

To find two values of $k$ such that the points $\left(-3,4\right),\left(0,k\right)\text{and}\left(k,10\right)$ are collinear.

Answer to Problem 34WE

The two values of $k$ are $-5,6$ .

Explanation of Solution

Given:

The points are $\left(-3,4\right),\left(0,k\right)\text{and}\left(k,10\right)$

Calculation:

Slope of a line when two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given is

  $m=\frac{y_2-y_1}{x_2-x_1}$

Points are $A\left(-3,4\right),B\left(0,k\right)\text{and }C\left(k,10\right)$

Since points are collinear so

  $\begin{array}{l}{m}_{AB}=m_{BC}=m_{AC}\\ \frac{k-4}{0-\left(-3\right)}=\frac{10-k}{k-0}=\frac{10-4}{k-\left(-3\right)}\end{array}$

Solving

  $\begin{array}{l}{m}_{AB}=m_{BC}\\ \frac{k-4}{0-\left(-3\right)}=\frac{10-k}{k-0}\\ \frac{k-4}{3}=\frac{10-k}{k}\\ k\left(k-4\right)=3\left(10-k\right)\\ k^2-4k=30-3k\\ k^2-k-30=0\\ k^2-6k+5k-30=0\\ k\left(k-6\right)+5\left(k-6\right)=0\\ \left(k-6\right)\left(k+5\right)=0\\ k-6=0\\ k=6\\ k+5=0\\ k=-5\end{array}$

Therefore, the two values of $k$ are $-5,6$ .