calculate the normal vector of line y = -0.75x + 3

The given line has a slope of -0.75. The slope of any line perpendicular to it would be the negative reciprocal of -0.75, which is the negative inverse of -0.75. Let's calculate the negative inverse of -0.75: Negative inverse of -0.75 = -1 / (-0.75) = 4/3 Therefore, the slope of the line perpendicular to y = -0.75x + 3 is 4/3 Now, we can express the normal vector in the form (a, b), where a and b represent the components of the vector. Since the slope is equal to the ratio of the vertical component to the horizontal component, we have: a / b = 4/3 By comparing the coefficients, we can see that a = 4 and b = 3. Therefore, the normal vector of the line y = -0.75x + 3 is \begin{bmatrix} 4 \\ 3 \end{bmatrix}