For this case we have to;
We have that an equation in slope-intercept form is given by:
[tex]y = mx + b[/tex]
Where:
m is the slope
b is the cut point with the y axis
Also, by definition, two lines are perpendicular when the product of their slopes is -1. That is:[tex]m_ {1} * m_ {2} = - 1[/tex]
We have the line as data: [tex]y_ {1} = \frac {-5} {2} x + 6[/tex]
Then [tex]m_ {1} = \frac {-5} {2}[/tex]
We found[tex]m_ {2}[/tex]:
[tex]m_ {1} * m_ {2} = - 1[/tex]
[tex]\frac {-5} {2} * m_ {2} = - 1[/tex]
[tex]m_ {2} = \frac {-1} {(\frac {-5} {2})}[/tex]
[tex]m_ {2} = \frac {(2) (- 1)} {(- 5) (1)}[/tex]
[tex]m_ {2} = \frac {2} {5}[/tex]
Thus, [tex]y_ {2} = \frac {2} {5} x_ {2} + b_ {2}[/tex]
We must find [tex]b_ {2}[/tex]:
We know that [tex]y_ {2}[/tex] passes through the point[tex](x_ {2}, y_ {2}) = (5,0)[/tex]
We substitute the point in the equation of [tex]y_ {2}[/tex]:
[tex]0 = \frac {2} {5} (5) + b_ {2}\\0 = 2 + b_ {2}\\b_ {2} = - 2[/tex]
Thus, [tex]y_ {2} = \frac {2} {5} x_ {2} -2[/tex]
Then the equation in slope-intercept for the line that passes through (5,0) and is perpendicular to the line described by [tex]y_ {1} = \frac {-5} {2} x_{1} + 6[/tex] is: [tex]y_ {2} = \frac {2} {5} x_ {2} -2[/tex]
Answer:
[tex]y_ {2} = \frac {2} {5} x_ {2} -2[/tex]
