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# Parrel lines i have no idea what this is

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Write an equation of the line that passes through (−4,−1)(−4,−1) and is perpendicular to the line y=43x+6

Guest Oct 30, 2017

#1
+7155
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The slope of the line  y = 43x + 6  is  43 .

The slope of a line perpendicular to that is  -1/43

So we want the equation of a line with a slope of  -1/43

that passes through  (-4, -1)  and a general point,  (x, y)

$$\text{slope}\,=\,\frac{y_2-y_1}{x_2-x_1} \\~\\ -\frac{1}{43}=\frac{y-(-1)}{x-(-4)} \\~\\ -\frac{1}{43}=\frac{y+1}{x+4} \\~\\ (-\frac{1}{43})(x+4)=y+1 \\~\\ -\frac1{43}x-\frac4{43}=y+1 \\~\\ -\frac1{43}x-\frac4{43}-\frac{43}{43}=y \\~\\ y=-\frac1{43}x-\frac{47}{43}$$

hectictar  Oct 30, 2017
#1
+7155
+2

The slope of the line  y = 43x + 6  is  43 .

The slope of a line perpendicular to that is  -1/43

So we want the equation of a line with a slope of  -1/43

that passes through  (-4, -1)  and a general point,  (x, y)

$$\text{slope}\,=\,\frac{y_2-y_1}{x_2-x_1} \\~\\ -\frac{1}{43}=\frac{y-(-1)}{x-(-4)} \\~\\ -\frac{1}{43}=\frac{y+1}{x+4} \\~\\ (-\frac{1}{43})(x+4)=y+1 \\~\\ -\frac1{43}x-\frac4{43}=y+1 \\~\\ -\frac1{43}x-\frac4{43}-\frac{43}{43}=y \\~\\ y=-\frac1{43}x-\frac{47}{43}$$

hectictar  Oct 30, 2017