For this case we have that by definition, the equation of the line in the slope-intersection form is given by:
![y = mx + b](https://img.qammunity.org/2020/formulas/mathematics/high-school/fc4cgm6covys37zv2opmmp9ps4jxyjepvh.png)
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:
![(x_ {1}, y_ {1}) :( 8, -10)\\(x_ {2}, y_ {2}): (- 8,10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/61v35j0dq881tpioovk03338nm651lo0p5.png)
We found the slope:
![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}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9dr13vv441rt17wsm3al8wzpf5vn0eimzi.png)
By definition, if two lines are perpendicular then the product of their slopes is -1:
![m * - \frac {5} {4} = - 1\\m = \frac {-1} {- \frac {5} {4}}\\m = \frac {4} {5}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/22np21djxq3b7qugwa8rkyj4frf21g9kvs.png)
Thus, the equation is of the form:
![y = \frac {4} {5} x + b](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ouibaki96kitalbzf85n9w438yffgem6nq.png)
If the line goes through
we have:
![3 = \frac {4} {5} (5) + b\\3 = 4 + b\\3-4 = b\\-1 = b](https://img.qammunity.org/2020/formulas/mathematics/middle-school/21pulgi5tfkc7266s6q4efj7xz1tyty7jc.png)
Finally, the equation is:
![y = \frac {4} {5} x-1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/szyx4qr3sgtmb6f0ngqafpowzor63ax38u.png)
Algebraically manipulating we have:
![y + 1 = \frac {4} {5} x\\5y + 5 = 4x\\4x-5y = 5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e8nclagit6t18hnagzmubsm9gizkm69nfm.png)
Answer:
Option A