What is the perpendicular y= 2/5 x+1,(10,-8) ?

The perpendicular y= 2/5 x+1,(10,-8) is y=-5/2 x+17

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perpendicular y= 2/5 x+1,(10,-8)

perpendicular

y = \frac{2}{5} x + 1, (10, - 8)

y = - \frac{5}{2} x + 17

Solution steps

Find the line

y = mx + b

perpendicular to

y = \frac{2}{5} x + 1

that passes through

(10, - 8)

Find the slope of

y = \frac{2}{5} x + 1 : m = \frac{2}{5}

Compute the slope of the perpendicular line:

m_{p} = - \frac{5}{2}

Find the line with slope m=

- \frac{5}{2}

and passing through

(10, - 8) : y = - \frac{5}{2} x + 17

y = - \frac{5}{2} x + 17

x^{2} + 5 x

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

f (x) = 3^{x - 2}

3 x + y = - 2

f (x) = e^{x^{2}} + ln (x^{2})

What is the perpendicular y= 2/5 x+1,(10,-8) ?
The perpendicular y= 2/5 x+1,(10,-8) is y=-5/2 x+17