What is the perpendicular y=8x-3,(6,2) ?

The perpendicular y=8x-3,(6,2) is y=-1/8 x+11/4

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perpendicular y=8x-3,(6,2)

perpendicular

y = 8 x - 3, (6, 2)

y = - \frac{1}{8} x + \frac{11}{4}

Solution steps

Find the line

y = mx + b

perpendicular to

y = 8 x - 3

that passes through

(6, 2)

Find the slope of

y = 8 x - 3 : m = 8

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{8}

Find the line with slope m=

- \frac{1}{8}

and passing through

(6, 2) : y = - \frac{1}{8} x + \frac{11}{4}

y = - \frac{1}{8} x + \frac{11}{4}

f (x) = \frac{\frac{x}{x + 7}}{\frac{x}{x + 7} + 7}

f (x) = \frac{5 x + 9}{6}

f (x) = ln (6 - x)

\frac{1}{x - 15}

f (x) = - 5 (x - 23)^{2} + 41

What is the perpendicular y=8x-3,(6,2) ?
The perpendicular y=8x-3,(6,2) is y=-1/8 x+11/4