For this case we have that by definition, the point-slope equation of a line is given by:
[tex]y-y_ {0} = m (x-x_ {0})[/tex]
Where:
m: It's the slope
[tex](x_ {0}, y_ {0})[/tex]: It is a point that belongs to the line
We find the slope with the given points:
[tex](x_ {1}, y_ {1}): (- 5,4)\\(x_ {2}, y_ {2}) :( 1,6)[/tex]
[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {6-4} {1 - (- 5)} = \frac {2} {1 +5} = \frac {2} {6} = \frac {1} {3}[/tex]
Then, the equation is of the form:
[tex]y-y_ {0} = \frac {1} {3} (x-x_ {0})[/tex]
We substitute the point [tex](1,6)[/tex]:
[tex]y-6 = \frac {1} {3} (x-1)[/tex]
Finally, the equation is:
[tex]y-6 = \frac {1} {3} (x-1)[/tex]
Answer:
[tex]y-6 = \frac {1} {3} (x-1)[/tex]
