## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.4 - Linear Functions and Slope - Exercise Set - Page 213: 21

#### Answer

point-slope form: $y=\frac{1}{2}(x)$ slope-intercept form: $y= \frac{1}{2}x$

#### Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where m = slope and b = y-intercept (2) The point-slope form of a line's equation is: $y-y_1=m(x-x_1)$ (a) point-slope form The given line has a slope of $\frac{1}{2}$ and passes through the origin, $(0, 0)$. Substitute these values into the point-slope form above to obtain: $y-0=\frac{1}{2}(x-0) \\y = \frac{1}{2}(x)$ (b) slope-intercept form Substitute the slope $\frac{1}{2}$ to $m$ to obtain the tentative equation: $y=\frac{1}{2}x+b$ The line passes through the origin, $(0, 0)$. This means that the coordinates of this point satisfy the equation of the line. Substitute the x and y-coordinates of this point into the tentative equation to obtain: $y=\frac{1}{2}x+b \\0 = \frac{1}{2}(0) + b \\0 = 0 + b \\0= b$ Thus, the equation of the line is $y=\frac{1}{2}x$.

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