What is the perpendicular y=0.3x+6,(3,-8) ?

The perpendicular y=0.3x+6,(3,-8) is y=-10/3 x+2

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perpendicular y=0.3x+6,(3,-8)

perpendicular

y = 0.3 x + 6, (3, - 8)

y = - \frac{10}{3} x + 2

Solution steps

Find the line

y = mx + b

perpendicular to

y = 0.3 x + 6

that passes through

(3, - 8)

Find the slope of

y = 0.3 x + 6 : m = 0.3

Compute the slope of the perpendicular line:

m_{p} = - \frac{10}{3}

Find the line with slope m=

- \frac{10}{3}

and passing through

(3, - 8) : y = - \frac{10}{3} x + 2

y = - \frac{10}{3} x + 2

f (x) = arctan (x)

f (x) = \frac{1}{x ^{2} - 2 x - x}

y = 2^{- x} + 1

lo g_{2} (2 - \frac{x}{3})

7 - x^{2}

What is the perpendicular y=0.3x+6,(3,-8) ?
The perpendicular y=0.3x+6,(3,-8) is y=-10/3 x+2