Question

Find the slope of the given line whose equation is \[3y = x + 3\].

Answer 1

Hint: In order to find the slope of the given line, we will first transform the equation of the line into the standard form of line which is \[y = mx + c\]where the value of m is equal to slope of the line. After transforming the equation we will compare the equation with the standard to determine slope.

Complete step by step solution:
Before dwelling into the question let us first understand what is the slope of an equation and how it is determined. Simply put, if theta is the inclination of any line L then tanθ is called the slope or gradient of the line L. The slope of a line is determined by m. Thus, m= tanθ, θ is not equal to 90°. It is important to note here that the slope of the x-axis is zero and the slope of y-axis is not defined.

Consequently, the slope of any line L with vertices \[\left( {{x_1},{y_1}} \right)\left( {{x_2},{y_2}} \right)\]is given by
\[ \Rightarrow m = \dfrac{{{y_2} - {y_1}}}{{{x_{2 - }}{x_1}}}\]
Now on the basis of our understanding of slopes we will try and solve our given question,
Given equation of line is:
\[ \Rightarrow 3y = x + 3\]
\[ \Rightarrow y = \dfrac{x}{3} + \dfrac{3}{3}\]
Gives, \[y = \dfrac{x}{3} + 1 - - - - - \left( 1 \right)\]
Now, we know that the equation of a line is given by,\[y = mx + c\]where the value of m gives the value slope.
 Here, comparing the equation (1) with the standard equation of line,
We get, \[m = \dfrac{1}{3}\]=slope of the given line.

Hence the correct answer is \[m = \dfrac{1}{3}\]

Note: It is very important to note here that two nonvertical lines are perpendicular to each other only if the product of their slopes is negative 1 or they are negative reciprocal of each other i.e., \[{m_2} = - \dfrac{1}{{{m_1}}}\] or \[{m_1}{m_2} = - 1\]. Also, two nonvertical lines are only parallel if their slopes are equal to each other i.e.,\[{m_1} = {m_2}\].