Question

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

Answer 1

nobody gonna see this but its A

Explanation:

trust me

Answer 2

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