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

Sagot :

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

[tex]y-y_1=m(x-x_1)[/tex]

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:

[tex]m=\frac{y_1-y_2_{}}{x_1-x_2}[/tex]

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)

[tex]\begin{gathered} m=\frac{6-3}{5-(-1)} \\ m=\frac{3}{5+1} \\ m=\frac{3}{6} \\ m=\frac{1}{3} \end{gathered}[/tex]

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:

[tex]y-6=\frac{1}{3}(x-5)[/tex]

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:

[tex]\begin{gathered} y-6=\frac{1}{3}\cdot x-\frac{1}{3}\cdot5 \\ y-6=\frac{1}{3}x-\frac{16}{3} \end{gathered}[/tex]

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

[tex]\begin{gathered} y-6+6=\frac{1}{3}x-\frac{16}{3}+6 \\ y=\frac{1}{3}x+\frac{2}{3} \end{gathered}[/tex]

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

[tex]y=\frac{1}{3}x+\frac{2}{3}[/tex]