Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.4 - Linear Functions and Slope - Exercise Set - Page 213: 23


Point-slope form is $y+2=-\frac{2}{3}\left( x-6 \right)$. Slope-intercept form is $y=-\frac{2}{3}x+2$.

Work Step by Step

The point slope form of a line is obtained with the help of the slope of the line and any one point that lies in the line. For a line with slope m and passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$, point-slope form is given as follows: $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Consider $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 6,-2 \right)$ Since the point-slope form is given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, replace the values of m and $\left( {{x}_{1}},{{y}_{1}} \right)$ in it: $\begin{align} & y-\left( -2 \right)=-\frac{2}{3}\left( x-6 \right) \\ & y+2=-\frac{2}{3}\left( x-6 \right) \\ & y+2=-\frac{2}{3}x+4 \end{align}$ Subtract $2$ from both sides of the equation: $\begin{align} & y=-\frac{2}{3}x+4-2 \\ & =-\frac{2}{3}x+2 \end{align}$ The point-slope form of line is $y=-\frac{2}{3}x+2$. The slope-intercept form of a line is provided by $y=mx+b$ ; here, m is the slope and b is the y-intercept, and the y-intercept is the y-coordinate of a point where the line intersects the y-axis. Change the values of m and $\left( x,y \right)=\left( 6,-2 \right)$ in $y=mx+b$ to find the value of b. $\begin{align} & -2=-\frac{2}{3}\left( 6 \right)+b \\ & b=2 \end{align}$ Now, change the values of b and m in $y=mx+b$. So $y=-\frac{2}{3}x+2$ The slope-intercept form of the equation is $y=-\frac{2}{3}x+2$. Therefore, the point-slope form of the line with $m=-\frac{2}{3}$ and passing through the point $\left( 6,-2 \right)$ is $y+2=-\frac{2}{3}\left( x-6 \right)$, and the slope-intercept form is $y=-\frac{2}{3}x+2$.
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