## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises - Page 298: 63

#### Answer

(a) slope-intercept form $\color{blue}{y=3x-7}$ (b) standard form $\color{red}{3x-y=7}$

#### Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. (2) The standard form of a line's equation is $Ax+By=C$ where $A\ge0$ and A, B, and C are integers. (3) The point-slope form of a line's equation is $y-y_1=m(x-x_1)$ where $m$=slope and $(x_1, y_1)$ is a point on the line. (4) The slope $m$ of the line that contains the points $(x_1. y_1)$ and $(x_2, y_2)$ is given by the formula: $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ (5) Parallel lines have the same slope. The line is parallel to $3x-y=1$. The slope-intercept form of this equation is $y=3x-1$. Since the slope of this line is $3$, then the slope of the line parallel to is also $3$. The line we are looking for has a slope of $3$ and contains the point $(2, -1)$. Thus, the point-slope form of the line's equation is: $$y-(-1)=3(x-2) \\y+1=3(x-2)$$ (a) slope-intercept form The slope-intercept form of the line can be derived from the equation above: $y+1=3(x-2) \\y+1=3(x)-3(2) \\y+1=3x-6 \\y+1-1=3x-6-1 \\\color{blue}{y=3x-7}$ (b) standard form The standard form of the line's equation can be derived from the slope-intercept form: $y=3x-7 \\y+7=3x-7+7 \\y+7=3x \\y+7-y=3x-y \\7=3x-y \\\color{red}{3x-y=7}$

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