What is the perpendicular 4x+3y-7=0,(6,-5) ?

The perpendicular 4x+3y-7=0,(6,-5) is y= 3/4 x-19/2

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perpendicular 4x+3y-7=0,(6,-5)

perpendicular

4 x + 3 y - 7 = 0, (6, - 5)

y = \frac{3}{4} x - \frac{19}{2}

Solution steps

Find the line

y = mx + b

perpendicular to

4 x + 3 y - 7 = 0

that passes through

(6, - 5)

Find the slope of

4 x + 3 y - 7 = 0 : m = - \frac{4}{3}

Compute the slope of the perpendicular line:

m_{p} = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(6, - 5) : y = \frac{3}{4} x - \frac{19}{2}

y = \frac{3}{4} x - \frac{19}{2}

f (x) = - 14 - x

f (x) = 3 lo g_{4} (x + 1) - 3

y = ln (x + 3) - 1

7 x (x + 1)

What is the perpendicular 4x+3y-7=0,(6,-5) ?
The perpendicular 4x+3y-7=0,(6,-5) is y= 3/4 x-19/2