Question

Write the slope-intercept form of the equation that passes through the point (0,-3) and is perpendicular to the line y = 2x - 6

Answer 1

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

$y = mx + b$

Where:

m: It's the slope

b: It is the cutoff point with the y axis

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

We have the following line:

$y = 2x-6$

Then$m_ {1} = 2$

The slope of a perpendicular line will be:

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

Thus, the equation of the line will be:

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

We substitute the given point and find "b":

$-3 = - \frac {1} {2} (0) + b\\-3 = b$

Finally the equation is:

$y = - \frac {1} {2} x-3$

Answer:

$y = - \frac {1} {2} x-3$

Answer 2

Answer:

$y=-\frac{1}{2}x -3$

Step-by-step explanation:

The slope-intercept form of the equation of a line has the following form:

$y=mx + b$

Where m is the slope of the line and b is the intercept with the y axis

In this case we look for the equation of a line that is perpendicular to the line

$y = 2x - 6$.

By definition If we have the equation of a line of slope m then the slope of a perpendicular line will have a slope of $-\frac{1}{m}$

In this case the slope of the line $y = 2x - 6$ is $m=2$:

Then the slope of the line sought is: $m=-\frac{1}{2}$

The intercept with the y axis is:

If we know a point $(x_1, y_1)$ belonging to the searched line, then the constant b is:

$b=y_1-mx_1$ in this case the poin is: (0,-3)

Then:

$b= -3 -(\frac{1}{2})(0)\\\\b=-3$

finally the equation of the line is:

$y=-\frac{1}{2}x-3$