Answer:
Part A)
The slope is 3.
Part B)
[tex]\displaystyle y-1=3(x+3)[/tex]
Part C)
[tex]\displaystyle y=3x+10[/tex]
Step-by-step explanation:
We know that the line goes through the points (-3, 1) and (-2, 4).
Part A)
To find the slope of a line given two points, we can use the slope formula:
[tex]\displaystyle m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Where (x₁, y₁) and (x₂, y₂) are our two points.
So, let’s let (-3, 1) be (x₁, y₁) and let (-2, 4) be (x₂, y₂). Substitute appropriately:
[tex]\displaystyle m=\frac{4-1}{-2-(-3)}[/tex]
Evaluate:
[tex]\displaystyle m=\frac{3}{-2+3}=\frac{3}{1}=3[/tex]
Hence, the slope of our line is 3.
Part B)
Point-slope form is given by:
[tex]\displaystyle y-y_1=m(x-x_1)[/tex]
Where m is our slope and (x₁, y₁) is a point.
So, let’s substitute 3 for m.
For our point, we can use either of the two given. Let’s use (-3, 1) for consistency. So, let (-3, 1) be (x₁, y₁). Therefore:
[tex]\displaystyle y-1=3(x-(-3))[/tex]
Simplify. Hence, our point-slope form is:
[tex]\displaystyle y-1=3(x+3)[/tex]
Part C)
To rewrite into slope-intercept form, we can just solve for y from our point-slope form. So, we have:
[tex]\displaystyle y-1=3(x+3)[/tex]
Distribute on the right:
[tex]\displaystyle y-1=3x+9[/tex]
Add 1 to boths sides. So, our slope-intercept equation is:
[tex]\displaystyle y=3x+10[/tex]
