[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-3+3}{1+2}\implies \cfrac{0}{3}\implies 0[/tex]
[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{0}[x-\stackrel{x_1}{(-2)}] \\\\\\ y+3=0\implies y=-3[/tex]
Answer:
Step-by-step explanation:
METHOD 1:
The sloope-intercept form of an equation of a line:
[tex]y=mx+b[/tex]
m - slope
b - y-intercept
The formula of a sloe:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
(x₁, y₁), (x₂, y₂) - points on a line
We have the points (-2, -3) and (1, -3).
Substitute:
[tex]m=\dfrac{-3-(-3)}{1-(-2)}=\dfrac{-3+3}{1+2}=\dfrac{0}{3}=0[/tex]
Put the value of the slope and the coordinates of the point (1, -3) to the equation of a line:
[tex]-3=0(1)+b\to b=-3[/tex]
Finally:
[tex]y=-3[/tex]
METHOD 2:
We can see that the second coordinates of the points (ordinate) are the same.
Conclusion: This is a horizontal line.
The equation of a horizontal line:
[tex]y=b[/tex]
We have (-2, -3), (1, -3) → y = -3
