Question

Which of the following represent a line parallel to the $x$–axis?
(a) $x + y = 3$ (b) $2x + 3 = 7$
(c) $2 - y - 3 = y + 1$ (d) $x + 3 = 0$

Answer 1

Here, we need to check which of the given lines is parallel to the $x$–axis. We will use the general form of a line parallel to the $x$–axis or the $y$–axis to check which of the given lines is parallel to the $x$–axis and find the correct option.

Complete step by step solution:
We will check each of the options one by one to find which of the given lines are parallel to the $x$–axis.
A line parallel to the $x$–axis is of the form $y + a = 0$, where $a$ is not equal to 0.
The first equation is $x + y = 3$.
Rewriting the equation $y + a = 0$, we get
$0 \cdot x + 1 \cdot y + a = 0$
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have unique solution if $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$.
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have infinitely many solutions if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have no solution if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$, and thus, are parallel.
Comparing $0 \cdot x + 1 \cdot y + a = 0$ to the standard form ${a_1}x + {b_1}y + {c_1} = 0$, we get
${a_1} = 0$, ${b_1} = 1$, and ${c_1} = a$
Comparing $x + y = 3$ to the standard form ${a_2}x + {b_2}y + {c_2} = 0$, we get
${a_2} = 1$, ${b_2} = 1$, and ${c_2} = - 3$
Now, we will find the ratios of the coefficients of $x$, $y$, and the constant.
Dividing ${a_1} = 0$ by ${a_2} = 1$, we get
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{0}{1} = 0$
Dividing ${b_1} = 1$ by ${b_2} = 1$, we get
$\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{1} = 1$
Therefore, we can observe that $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$.
Therefore, the lines $0 \cdot x + 1 \cdot y + a = 0$ and $x + y = 3$ have a unique solution.
This means that these lines are intersecting lines.
Hence, $x + y = 3$ is not parallel to the $x$–axis.
Thus, option (a) is incorrect.
The second equation is $2x + 3 = 7$.
Subtracting 3 from both sides, we get
$\begin{array}{l} \Rightarrow 2x + 3 - 3 = 7 - 3\\\ \Rightarrow 2x = 4\end{array}$
Dividing both sides by 2, we get
$\Rightarrow x = 2$
Rewriting the equation, we get
$\Rightarrow x - 2 = 0$
We can observe that this equation is of the form $x + a = 0$.
A line parallel to the $y$–axis is of the form $x + a = 0$, where $a$ is not equal to 0.
Thus, the line $2x + 3 = 7$ is parallel to the $y$–axis.
Therefore, the line $2x + 3 = 7$ is not parallel to the $x$–axis.
Thus, option (b) is incorrect.
The third equation is $2 - y - 3 = y + 1$.
Adding and subtracting the like terms, we get
$\Rightarrow - 1 - y = y + 1$
Rewriting the equation, we get
$\Rightarrow y + y = - 1 - 1$
Simplifying the expression, we get
$\Rightarrow 2y = - 2$
Dividing both sides by 2, we get
$\Rightarrow y = - 1$
Rewriting the equation, we get
$\Rightarrow y + 1 = 0$
We can observe that this equation is of the form $y + a = 0$.
A line parallel to the $x$–axis is of the form $y + a = 0$, where $a$ is not equal to 0.
Therefore, the line $2 - y - 3 = y + 1$ is parallel to the $x$–axis.
Thus, option (c) is the correct option.
The fourth equation is $x + 3 = 0$.
We can observe that this equation is of the form $x + a = 0$.
A line parallel to the $y$–axis is of the form $x + a = 0$, where $a$ is not equal to 0.
Thus, the line $x + 3 = 0$ is parallel to the $y$–axis.
Therefore, the line $x + 3 = 0$ is not parallel to the $x$–axis.

Note:
A line parallel to the $x$–axis is of the form $y + a = 0$, where $a$ is not equal to 0. If $a$ is equal to 0, then the line becomes $y = 0$, which is coincident with the $x$–axis, and not parallel. Hence, $a$ cannot be equal to 0. Similarly, a line parallel to the $y$–axis is of the form $x + a = 0$, where $a$ is not equal to 0.