A line passes through the origin and the point (-6, 4).

Write the equation of the line that passes through (-6, 4) and is perpendicular to the given line.

Write the final answer in the slope-intercept form y = mx + b.
(The slope and y-intercept must be written as fractions, when needed).

Y=-6x+10

To find the equation of the line that is perpendicular to the given line, we first need to find the slope of the given line.

The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:

slope = (y2 - y1) / (x2 - x1)

In this case, the given line passes through the origin (0, 0) and the point (-6, 4). So, substituting the coordinates in the above formula, we have:

slope = (4 - 0) / (-6 - 0)
= 4 / -6
= -2/3

Since the line we want to find is perpendicular to the given line, the slope of the new line will be the negative reciprocal of the given slope. So the slope of the new line is the negative reciprocal of -2/3.

The negative reciprocal of a fraction m is given by -1/m. So, the negative reciprocal of -2/3 is:

-1 / (-2/3) = -1 * (-3/2) = 3/2

Now, we have the slope of the new line, which is 3/2. We also have one point (-6, 4) that the line passes through.

Using the point-slope form of the equation of a line, which is:

y - y1 = m(x - x1),

we can substitute the values into the equation to get:

y - 4 = (3/2)(x - (-6))
y - 4 = (3/2)(x + 6)
y - 4 = (3/2)x + 9
y = (3/2)x + 9 + 4
y = (3/2)x + 13.

Therefore, the equation of the line that passes through (-6, 4) and is perpendicular to the given line is y = (3/2)x + 13.