Question

Write an equation of the line that passes through the points (-2,-3) and (1-3)

Answer 1

Answer:

y = -3

Step-by-step explanation:

METHOD 1:

The sloope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a sloe:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

(x₁, y₁), (x₂, y₂) - points on a line

We have the points (-2, -3) and (1, -3).

Substitute:

$m=\dfrac{-3-(-3)}{1-(-2)}=\dfrac{-3+3}{1+2}=\dfrac{0}{3}=0$

Put the value of the slope and the coordinates of the point (1, -3) to the equation of a line:

$-3=0(1)+b\to b=-3$

Finally:

$y=-3$

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:

$y=b$

We have (-2, -3), (1, -3) → y = -3

Answer 2

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-3+3}{1+2}\implies \cfrac{0}{3}\implies 0$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{0}[x-\stackrel{x_1}{(-2)}] \\\\\\ y+3=0\implies y=-3$