Find the equation of a line perpendicular to4, x, minus, 3, y, equals, 124x−3y=12that passes through the point left bracket, 0, comma, minus, 3, right bracket(0,−3).

First, let's find the slope of the given line. The equation of the line is in the form Ax + By = C, which can be rewritten in the slope-intercept form (y = mx + b) as y = -A/Bx + C/B.

For the given line 4x - 3y = 12, A = 4 and B = -3. So, the slope of the given line (m1) is -A/B = -4/-3 = 4/3.

The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. So, the slope of the line we are trying to find (m2) is -1/m1 = -1/(4/3) = -3/4.

Now, we know that the line we are looking for has a slope of -3/4 and passes through the point (0, -3). The equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Since the line passes through (0, -3), the y-intercept is -3.

So, the equation of the line is y = -3/4x - 3.