Question

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

Answer 1

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 slope

b: It is the cut-off point with the y axis

According to the figure, the line goes through the following points:

$(x_ {1}, y_ {1}) :( 8, -10)\\(x_ {2}, y_ {2}): (- 8,10)$

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