What is the perpendicular 2x-7y+10=0,(7,1) ?

The perpendicular 2x-7y+10=0,(7,1) is y=-7/2 x+51/2

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perpendicular 2x-7y+10=0,(7,1)

perpendicular

2 x - 7 y + 10 = 0, (7, 1)

y = - \frac{7}{2} x + \frac{51}{2}

Solution steps

Find the line

y = mx + b

perpendicular to

2 x - 7 y + 10 = 0

that passes through

(7, 1)

Find the slope of

2 x - 7 y + 10 = 0 : m = \frac{2}{7}

Compute the slope of the perpendicular line:

m_{p} = - \frac{7}{2}

Find the line with slope m=

- \frac{7}{2}

and passing through

(7, 1) : y = - \frac{7}{2} x + \frac{51}{2}

y = - \frac{7}{2} x + \frac{51}{2}

r an g e f (x) = e^{x - 1}

r an g e f (x) = - 4

in f l ec t i o n x^{3} - x

in v erse y = \frac{( x + 1 )}{x - 3}

p er i o d i c i t y 300 sin (7 t + π)

What is the perpendicular 2x-7y+10=0,(7,1) ?
The perpendicular 2x-7y+10=0,(7,1) is y=-7/2 x+51/2