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: 25


Point-slope form is $y-2=2\left( x-1 \right)$. Slope-intercept form is $y=2x$.

Work Step by Step

The point slope form of a line can be obtained with the help of the slope of the line and any one point that lies on the line. For a line with slope m and passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ , the point-slope form is given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Consider $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,2 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 5,10 \right)$ At first, calculate the slope of the line by $m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ ; here, ${{x}_{2}}-{{x}_{1}}\ne 0$ Substitute the value of $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in it: $\begin{align} & m=\frac{10-2}{5-1} \\ & =\frac{8}{4} \\ & =2 \end{align}$ The point-slope form is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. Replace the values of m and $\left( {{x}_{1}},{{y}_{1}} \right)$ in the above equation to get $\begin{align} & y-2=2\left( x-1 \right) \\ & y-2=2x-2 \\ & y=2x-2+2 \\ & =2x \end{align}$ The point-slope form of the equation is $y=2x$. The slope-intercept form of the line is given 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. Substitute the value of m and take the point $\left( x,y \right)=\left( 1,2 \right)$ in $y=mx+b$ to get the value of b: $\begin{align} & 2=2\left( 1 \right)+b \\ & b=0 \end{align}$ Now, replace the values of b and m in $y=mx+b$. Thus $y=2x+0$ The slope-intercept form of the line is $y=2x$. Hence, the point-slope form of line passing through the points $\left( 1,2 \right)$ and $\left( 5,10 \right)$ is $y-2=2\left( x-1 \right)$ , and the slope-intercept form is $y=2x$.
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