Question

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

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Cristian

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104 Answers

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}

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