Chapter 4.2, Problem 24PPS

To determine

Write an equation in point-slope form, slope-intercept form, and standard form for each line.

Answer to Problem 24PPS

Point-slope form of line is $y-2=\frac{17}{10}(x-1)$ .

The slope- intercept form of line is $y=\frac{17}{10}x+\frac{3}{10}$ .

The standard form of line is $-17x+10y=3$ .

Explanation of Solution

Given information: Slope of line $(m)=\frac{17}{10}$

A point on the line is ( x, y ) = $(1,2)$

Formulas used: i) The point-slope form of line is $y-y_1=m(x-x_1),$ where m is slope and $(x_1,y_1)$ is coordinate of a point on the line.

ii) Slope- intercepts form:

  $y=mx+b$ , where m is slope and b is y -intercept.

iii) Standard form

  $ax+by=c.$

Solution: Plugging the value of m = $\frac{17}{10}$ and $(x_1,y_1)=(1,2)$ in $y-y_1=m(x-x_1)$ .

$y-2=\frac{17}{10}(x-1)$

Therefore, the point-slope form of line is $y-2=\frac{17}{10}(x-1)$ .

For slope-intercept form of line we need to simplify the above equation.

$y-2=\frac{17}{10}(x-1)$ (Distribute $\frac{17}{10}$ )

$y-2=\frac{17}{10}x-\frac{17}{10}$

Add 2 on both sides

$y-2+2=\frac{17}{10}x-\frac{17}{10}+2$

$\Rightarrow y=\frac{17}{10}x+\frac{3}{10}$

Therefore, the slope- intercept form of line is $y=\frac{17}{10}x+\frac{3}{10}$ .

Let us find the standard form of line now.

Multiply each term by 10

$\begin{array}{l}10\cdot y=10\cdot \frac{17}{10}x+10\cdot \frac{3}{10}\\ \Rightarrow 10y=17x+3\end{array}$

Subtract 17x on both sides

$\begin{array}{l}-17x+10y=17x-17x+3\\ \Rightarrow -17x+10y=3\end{array}$

Therefore, the standard form of line is $-17x+10y=3$ .