Question

How do you find the slope, point-slope, slope-intercept, standard form, domain and range for line A(0,-7) (0,-12)?

Answer 1

This type of question is based on the concept of equation of lines. We have to find the slope with the given points, that is $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Here, we find that ${{x}_{1}}=0$, ${{y}_{1}}=-7$, ${{x}_{2}}=0$ and ${{y}_{2}}=-12$. Substitute the values in the formula for slope and do necessary calculations. We find that the denominator of the slope is 0 which means the slope is undefined. Thus, the equation of the line is x=0. Since, the slope is undefined we cannot find the point-slope and slope-intercept form of the equation. We know that the standard form of an equation is Ax+By=C. from the observations above, we get A=1, B=0 and C=0. On substituting the values of A, B and C, we get the standard form of the equation. Since x=0, the domain of the equation will be {0} and since the line x=0 extends to infinity the range will be $\left( -\infty ,\infty \right)$.

Complete step by step solution:
According to the question, we are asked to find the slope, point-slope, slope-intercept, standard form, domain and range for line A(0,-7) (0,-12).
We have been given the points are (0,-7) and (0,-12).
Let us assume (0,-7) to be $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ to be (0,-12).
We get ${{x}_{1}}=0$, ${{y}_{1}}=-7$, ${{x}_{2}}=0$ and ${{y}_{2}}=-12$.
We know that the formula to find the slope when two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are given is
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Let us now substitute the values in the formula.
$\Rightarrow m=\dfrac{-12-\left( -7 \right)}{0-0}$
On further simplification, we get
$m=\dfrac{-12+7}{0-0}$
$\Rightarrow m=\dfrac{-5}{0-0}$
$\therefore m=\dfrac{-5}{0}$
But we know that any term divided by 0 is undefined.
Therefore, the slope m is undefined.
Whenever we get the slope as undefined, the equation of the line will be x=0.
Let us find the point-intercept form of the equation.
We know that the point-slope form of an equation is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
Let us substitute the values in the equation.
$\Rightarrow y-\left( -7 \right)=\dfrac{-5}{0}\left( x-0 \right)$
On further simplification, we get
$y+7=\dfrac{-5}{0}x$
Here, we find that the denominator of the RHS has 0.
Therefore, the point-slope form of the equation cannot be determined.
Let us now find the slope- intercept form.
We know that the point-slope form of an equation is $y=mx+c$, where c is the intercept of the equation.
Let us substitute the values in the equation.
$\Rightarrow y=\dfrac{-5}{0}x+c$
Here, we find that the denominator of the RHS has 0.
Therefore, the slope-intercept form of the equation cannot be determined.
Now, let us find the standard form of the equation of line.
We know that the standard form of a line is Ax+By=C.
But we have found that the equation of line is x=0.
Therefore, by comparison we get
$x+\left( 0 \right)y=0$
That is A=1, B=0 and C=0.
Thus, the standard form of the line is $x+\left( 0 \right)y=0$.
We know that the domain of an equation is the set of all possible values of x for which the equation is determined.
Here, we find that x=0.
Therefore, the domain of the equation is {0}.
Range of a function is the set of all possible values of the dependent variable.
Here, the line x=0 extends to infinity on both the sides, that is from $-\infty$ to $\infty$.
Therefore, the range of the equation is $\left( -\infty ,\infty \right)$.
Hence, slope of the line A(0,-7) (0,-12) is undefined. The point-slope form and the slope-intercept form of the line cannot be found. The standard equation of the line is $x+\left( 0 \right)y=0$. The domain and the range is {0} and $\left( -\infty ,\infty \right)$ respectively.

Note: We should know that, when the slope is undefined the equation of the line is x=0 and not y=0. When we write the domain of the function, we have to express them as a set with a flower bracket. Also, the range should not be written in a closed bracket, that is $\left[ -\infty ,\infty \right]$.