What is the perpendicular 4x+7y-9=0,(5,3) ?

The perpendicular 4x+7y-9=0,(5,3) is y= 7/4 x-23/4

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perpendicular 4x+7y-9=0,(5,3)

perpendicular

4 x + 7 y - 9 = 0, (5, 3)

y = \frac{7}{4} x - \frac{23}{4}

Solution steps

Find the line

y = mx + b

perpendicular to

4 x + 7 y - 9 = 0

that passes through

(5, 3)

Find the slope of

4 x + 7 y - 9 = 0 : m = - \frac{4}{7}

Compute the slope of the perpendicular line:

m_{p} = \frac{7}{4}

Find the line with slope m=

\frac{7}{4}

and passing through

(5, 3) : y = \frac{7}{4} x - \frac{23}{4}

y = \frac{7}{4} x - \frac{23}{4}

f (x) = \frac{64}{x ^{2}}

12 x^{2} - 72 x + 77

y = x^{2} - 8 x - 7

f (x) = sin^{2} (15 x)

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

What is the perpendicular 4x+7y-9=0,(5,3) ?
The perpendicular 4x+7y-9=0,(5,3) is y= 7/4 x-23/4