Chapter 9, Problem 17CR

To determine

To calculate: Thevalue of k.

Answer to Problem 17CR

The value of k is $27$ .

Explanation of Solution

Given information:

Points at $(1,3),(-4,7),$ and $(-29,k)$ are collinear.

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)$ =coordinates of first point in the line.

  $(x_2,y_2)$ =coordinates of second point in the line.

Equations of straight lines are in the form $y=mx+c$ (m and c are numbers).m is the gradient of the line and c is the y-intercept.

Calculation:

Points are said to be collinear if they lie on straight line.

For $(1,3)$ and $(-4,7)$

A slope would be given by the following expression:

  $m=slope=\frac{y_2-y_1}{x_2-x_1}=\frac{7-3}{-4-1}=-\frac{4}{5}$

The equation of straight line is in the form as follows:

  $y=mx+c$

Points at $(1,3)$ are collinear.

  $\begin{array}{l}3=-\frac{4}{5}(1)+b\\ \frac{19}{5}=b\\ \therefore b=\frac{19}{5}\end{array}$

The equation of straight line is in the form as follows:

  $y=mx+c$

Points at $(-29,k)$ are collinear.

Thus,

  $\begin{array}{l}k=-\frac{4}{5}(-29)+\frac{19}{5}\\ =\frac{116}{5}+\frac{19}{5}\\ =\frac{135}{5}\\ =27\end{array}$