Question

PLEASE HELP!!! Write the equation of a line that goes through the point (6, 3) and is perpendicular to the line y = 4x + 1.

y−3=4(x−6)

y−3=−14(x−6)

y−6=4(x−3)

y−6=−14(x−3)

Answer 1

For this case we have that by definition, the equation of a line in the point-slope form is given by:

`y-y_ {0} = m (x-x_ {0})`

Where:

m: It is the slope of the line and
`(x_ {0}, y_ {0})`is a point through which the line passes.

We have the following equation of the slope-intersection form:

`y = 4x + 1`

Where the slope is
`m = 4`

By definition, if two lines are perpendicular then the product of their slopes is -1.

Thus, a perpendicular line will have a slope:

`m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {4}\\m_ {2} = - \frac {1} {4}`

Thus, the equation will be of the form:

`y-y_ {0} = - \frac {1} {4} (x-x_ {0})`

Finally we substitute the given point and we have:

`y-3 = - \frac {1} {4} (x-6)`

Answer:

Option B