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**+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

#1**+2 **

Best Answer

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