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

Last updated date: 01st Mar 2024
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Hint: 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.

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 $\left( -5,-1 \right)$.
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$
Let us take ${{x}_{1}}=-3$, ${{y}_{1}}=-1$, ${{x}_{2}}=-5$, ${{y}_{2}}=-1$ and substitute these in the above equation to get
\begin{align} & \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{align}
Subtracting $1$ from both the sides, we get
\begin{align} & \Rightarrow y+1-1=0-1 \\ & \Rightarrow y=-1 \\ & \Rightarrow y=\left( 0 \right)x-1 \\ \end{align}
Comparing the point slope form of the line, which is given by $y=mx+c$, we get $m=0$ and $c=-1$.
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 $\left( 0,-1 \right)$.