Question

How do you find the slope given $\left( -3,-1 \right)$ and $\left( -5,-1 \right)$?

Answer 1

From the coordinates of two points given in the question, which are $\left( -3,-1 \right)$ and $\left( -5,-1 \right)$, we can find the equation of the line passing through them. For this we need to use the two point form of the equation of line which is given as $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$. From the coordinates given in the question, we can choose ${{x}_{1}}=-3$, ${{y}_{1}}=-1$, ${{x}_{2}}=-5$, ${{y}_{2}}=-1$ and substitute it in the two point equation to get the equation of the line. Then, comparing the obtained equation with the point slope form given by $y=mx+c$ we can determine the required value of the slope.

Complete step by step answer:
Let us label the two points given in the question as A and B so that the coordinates of point A are $\left( -3,-1 \right)$ and that of the point B are $\left( -5,-1 \right)$.
Now, since we are given information about the two points, we can obtain the equation of line from them. This is because we know that the only curve which can be completely described by two points is a line. We have the coordinates of two points, so we apply the two point form to get the equation of the line passing through A and B. The two point form of a line, as we know, is given by
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$
Let us take ${{x}_{1}}=-3$, ${{y}_{1}}=-1$, ${{x}_{2}}=-5$, ${{y}_{2}}=-1$ and substitute these in the above equation to get

& \Rightarrow y-\left( -1 \right)=\dfrac{-1-\left( -1 \right)}{-5-\left( -3 \right)}\left( x-\left( -3 \right) \right) \\\ & \Rightarrow y+1=\dfrac{-1+1}{-5+3}\left( x+3 \right) \\\ & \Rightarrow y+1=\dfrac{0}{-2}\left( x+3 \right) \\\ & \Rightarrow y+1=0 \\\ \end{aligned}$$ Subtracting $1$ from both the sides, we get $\begin{aligned} & \Rightarrow y+1-1=0-1 \\\ & \Rightarrow y=-1 \\\ & \Rightarrow y=\left( 0 \right)x-1 \\\ \end{aligned}$ Comparing the point slope form of the line, which is given by $y=mx+c$, we get $$m=0$$ and $$c=-1$$. **Hence, the slope is equal to zero.** **Note:** From the coordinates given in the question, we can observe that the y-coordinate for both the points is the same each being equal to $-1$. Interpreting this information geometrically, we can appreciate that the line passing through both the points will be a horizontal line cutting the y-axis at $\left( 0,-1 \right)$.