Write the equation of the line containing (-3,4) and perpendicular to y = 2/3x + 2.

Write your answer in slope-intercept form.

Not sure how to do this one.

please help.

To find the equation of a line perpendicular to another line, you need to find the negative reciprocal of the slope of the original line.

The equation y = 2/3x + 2 is in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

The slope of the original line is 2/3. To find the negative reciprocal, you take the reciprocal, which means flipping the fraction upside down, and then take the negative sign:

Negative reciprocal = -3/2

So, the slope of the line perpendicular to y = 2/3x + 2 is -3/2.

Now, to find the equation of the line containing the point (-3,4) and having the slope -3/2, you can use the point-slope form of a line:

y - y1 = m(x - x1)

where (x1, y1) represents the coordinates of the given point, and m represents the slope.

Plugging in the values (-3,4) for (x1,y1) and -3/2 for m, we get:

y - 4 = -3/2(x - (-3))

Simplifying further:

y - 4 = -3/2(x + 3)

Expanding the brackets:

y - 4 = -3/2x - 9/2

Now, rearranging the equation to slope-intercept form, we isolate y on one side:

y = -3/2x - 9/2 + 4

Simplifying:

y = -3/2x - 9/2 + 8/2

y = -3/2x - 1/2

Therefore, the equation of the line perpendicular to y = 2/3x + 2 and containing the point (-3,4) is y = -3/2x - 1/2, in slope-intercept form.