Chapter 4.3, Problem 11PPS

To determine

To find: The equation of the straight line in point-slope form which passes through the point $\left(5,3\right)$ having slope $\left(m\right)=7$ . And, then graph the equation of the straight line.

Answer to Problem 11PPS

The equation of straight line in point-slope form is $y-3=7\left(x-5\right)$ .

Explanation of Solution

Given information: The slope of the given line is $\left(m\right)=7$ and it passes through the point $\left(4,3\right)$ .

Formula used: The equation of the straight line in point-slope form which passes through the points $\left(x_1,y_1\right)$ and having slope $\left(m\right)$ is written as $y-y_1=m\left(x-x_1\right)$ .

Calculation: The equation of the straight line in point-slope form having slope $\left(m\right)=7$ and passing through the point $\left(5,3\right)$ is written as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\begin{array}{l}\text{ Here, }m=7\\ \text{and }\left(x_1,y_1\right)\to \left(5,3\right)\end{array}\right)\\ y-3=7\left(x-5\right)\end{array}$

Graph: The table for the straight line $y-3=7\left(x-5\right)$ for respective values of $x$ and $y$ is shown as,

x	0	4.571
y	-32	0

Now, joining the points $\left(0,-32\right)$ and $\left(4.571,0\right)$ , the obtained graph is shown as

  

Interpretation: The obtained function is a straight line passing through the points $\left(0,-32\right)$ and $\left(4.571,0\right)$ .