I don’t understand this question what so ever
1 answer:
bearing in mind that perpendicular lines have <u>negative reciprocal</u> slopes, let's find firstly the slope of AC.
![\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-1}{1-2}\implies \cfrac{5}{-1}\implies -5 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-5}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{5}}}](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B6-1%7D%7B1-2%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B-1%7D%5Cimplies%20-5%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B%5Ccfrac%7B-5%7D%7B1%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B%5Ccfrac%7B1%7D%7B-5%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B%5Ccfrac%7B1%7D%7B5%7D%7D%7D)
so, we're really looking for the equation of a line whose slope is 1/5 and that passes through (3,3)

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Answer:
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Step-by-step explanation:
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