For this case we have that by definition, the equation of the line in the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cut-off point with the y axis
According to the figure, the line goes through the following points:
[tex](x_ {1}, y_ {1}) :( 8, -10)\\(x_ {2}, y_ {2}): (- 8,10)[/tex]
We found the slope:
[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {10 - (- 10)} {- 8-8} = \frac {10 + 10 } {- 16} = \frac {20} {- 16} = - \frac {5} {4}[/tex]
By definition, if two lines are perpendicular then the product of their slopes is -1:
[tex]m * - \frac {5} {4} = - 1\\m = \frac {-1} {- \frac {5} {4}}\\m = \frac {4} {5}[/tex]
Thus, the equation is of the form:
[tex]y = \frac {4} {5} x + b[/tex]
If the line goes through [tex](x, y) :( 5,3)[/tex]we have:
[tex]3 = \frac {4} {5} (5) + b\\3 = 4 + b\\3-4 = b\\-1 = b[/tex]
Finally, the equation is:
[tex]y = \frac {4} {5} x-1[/tex]
Algebraically manipulating we have:
[tex]y + 1 = \frac {4} {5} x\\5y + 5 = 4x\\4x-5y = 5[/tex]
Answer:
Option A
