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what is the equation of the line passing through the point (4, -2) and perpendicular to y =1/3x - 7
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3y + 6 = x-4 3y = x -10
For this, you have to know the good-ole point-slope form of a line:\[y-y_{0}=m(x-x_{0}),\]where\[y_{0}\]and\[x_{0}\]are the coefficients of the point you're given, and m is the slope of the line. Well, you're given a point as well as a slope (you know that the line should be perpendicular to y=1/3x-7). Therefore, your new slope should be m=3. Can you plug in these values into the point-slope equation? That is:\[y_{0}=-2\]\[x_{0}=4\]\[m=3\]into\[y-y_{0}=m(x-x_{0})?\]
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