Question

Write the equation of the line that is perpendicular to the line y = 2x + 2 and passes through the point (6, 3).

Answer 1

For this case we have that by definition, the equation of the line in a 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

We have the following equation:

$y = 2x + 2$

Thus, the slope is $m_ {1} = 2$

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

$m_ {1} * m_ {2} = - 1$

We find$m_ {2}:$

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {2}$

Thus, the equation of the line is:

$y = - \frac {1} {2} x + b$

We substitute the given point to find "b":

$3 = - \frac {1} {2} 6 + b\\3 = -3 + b\\b = 6$

Thus, the equation of a line perpendicular to the given line and passing through the given point is:

$y = - \frac {1} {2} x + 6$

Answer:

$y = - \frac {1} {2} x + 6$

Answer 2

Answer:

Step-by-step explanation: