Question

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

Answer 1

y = -3

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=(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=(-3-(-3))/(1-(-2))=(-3+3)/(1+2)=(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}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`