Find the equation of the line perpendicular to 3x-2y=3 and passing through (-3,-7)?

To find the equation of the line perpendicular to 3x-2y=3, we first need to find the slope of the given line.

Rearranging the equation 3x-2y=3 into slope-intercept form (y=mx+b), we get:
-2y = -3x + 3
y = (3/2)x - 3/2

The slope of the given line is 3/2.

Since the line we are looking for is perpendicular to the given line, the slope of the perpendicular line will be the negative reciprocal of 3/2, which is -2/3.

Now we have the slope (-2/3) and a point (-3,-7) that the line passes through. We can use the point-slope form of the equation of a line to find the equation of the perpendicular line:

y - y1 = m(x - x1)
y + 7 = (-2/3)(x + 3)
y + 7 = (-2/3)x - 2
y = (-2/3)x - 9

Therefore, the equation of the line perpendicular to 3x-2y=3 and passing through (-3,-7) is y = (-2/3)x - 9.

The equation of the given line is 3x-2y=3. The slope of this line is 3/2. The slope of the line perpendicular to this line is -2/3. The equation of the line perpendicular to 3x-2y=3 and passing through (-3,-7) is:

y - (-7) = -2/3(x - (-3))
y + 7 = -2/3(x + 3)
3(y + 7) = -2(x + 3)
3y + 21 = -2x - 6
2x + 3y = -27