Perpendicular to 3x-2y=3 and passing through (3, -7)
To find a line perpendicular to the given line 3x-2y=3, we first need to find the slope of the given line. We can rewrite the equation in slope-intercept form, y=mx+b, where m is the slope:
3x - 2y = 3
-2y = -3x + 3
y = (3/2)x - 3/2
So, the slope of the given line is 3/2.
The slope of a line perpendicular to this line will be the negative reciprocal of 3/2, which is -2/3.
Now, we can use the point-slope form of the equation of a line to find the equation of the line passing through (3, -7) with a slope of -2/3:
y - (-7) = -2/3(x - 3)
y + 7 = -2/3(x - 3)
y + 7 = -2/3x + 2
y = -2/3x - 5
Therefore, the equation of the line perpendicular to 3x-2y=3 and passing through (3, -7) is y = -2/3x - 5.