32. Persamaan garis yang melalui titik (8,- (2) dan tegak lurus garis dengan gradien (3)/(4) adalah __

The equation of the line is y = -(4/3)x + b

The problem relates to the field of analytical geometry in Mathematics. It is asking for the linear equation of a line that passes through the point (8, -2) and is perpendicular to a line with a gradient (slope) of 3/4. In linear equations, the slope of a line perpendicular to another is the negative reciprocal of the original line's slope. Therefore, for a line perpendicular to one with a gradient of 3/4, our perpendicular line will have the slope of -(4/3). Similarly, any line can be represented by the equation y = mx + b, where m represents the slope and b represents the y-intercept. However, since we have the point (8, -2) that the line must pass through, we substitute these values into our equation to get the specific equation of our required line. Hence, -2 = (-4/3)*8 + b. After solving this equation for b, we get a specific value. The equation of our line is then y = -(4/3)x + b.