What is the equation of the line that is parallel to the line y − 1 = 4(x + 3) and passes through the point (4, 32)?

y = –one-fourthx + 33
y = –one-fourthx + 36
y = 4x − 16
y = 4x + 16

Option D

The equation of the line that is parallel to the line y − 1 = 4(x + 3) and passes through the point (4, 32) is y = 4x + 16

Solution:

Given that the line passes through point (4, 32)

The line is parallel to y − 1 = 4(x + 3) ---- eqn 1

First let us find the slope of line

The general form of point slope form of line is given as:

 ----- eqn 2

Where "m" is the slope of line

Comparing the eqn 2 with eqn1  y − 1 = 4(x + 3)

We get "m" = 4

The slopes of parallel lines are always equal

Hence the slope of line passing through (4, 32) has the same slope "m" = 4

Now plug in "m" = 4 and in eqn 2 to get the required equation of line

y - 32 = 4(x - 4)

y - 32 = 4x - 16

y = 4x - 16 + 32

y = 4x + 16

Thus the equation of line parallel to given line is y = 4x + 16

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x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)