Q:

What is the equation of the line that is perpendicular to the given line and passes through the point (5, 3)

Accepted Solution

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