Question

Find an equation for the line that passes through the points (-1,3) and (5,6)

Answer 1

To determine the equation of a line we can use the point-slope form:

`y-y_1=m(x-x_1)`

Where

m is the slope of the line

(x₁,y₁) are the coordinates of one point on the line.

Since we know two points that are crossed by the line (5,6) and (-1,3) we can calculate the slope of the line using the following formula:

`m=\frac{y_1-y_2_{}}{x_1-x_2}`

Where

(x₁,y₁) are the coordinates of one point of the line

(x₂,y₂) are the coordinates of a second point of the line

Replace the formula using

(x₁,y₁) = (5,6)

(x₂,y₂) = (-1,3)

`\begin{gathered} m=(6-3)/(5-(-1)) \\ m=(3)/(5+1) \\ m=(3)/(6) \\ m=(1)/(3) \end{gathered}`

The slope of the line we are looking for is m=1/3

Next is to replace the coordinates of one of the points, for example (5,6) and the slope m=1/3 in the point-slope form to determine the equation of the line:

`y-6=(1)/(3)(x-5)`

If you need to express the equation in the slope-intercept form you can dos as follows:

-First, distribute the multiplication on the parentheses term:

`\begin{gathered} y-6=(1)/(3)\cdot x-(1)/(3)\cdot5 \\ y-6=(1)/(3)x-(16)/(3) \end{gathered}`

-Second, pass "-6" to the other side of the equation by applying the opposite operation, "+6" to both sides of it:

`\begin{gathered} y-6+6=(1)/(3)x-(16)/(3)+6 \\ y=(1)/(3)x+(2)/(3) \end{gathered}`

So the equation of the line that passes through points (-1,3) and (5,6) is

`y=(1)/(3)x+(2)/(3)`