Question

Find the equation of the line that is perpendicular to the line formed by the equation 3x + 2y = -14 and passes through the point (-48, 3).

Answer 1

First make this in y=mx+b form

Answer 2

Then find the slope (m) and make it negative for the perpendicular line

then, use point slope form with that new slope and build your new line equation

y-y1 = m(x-x1)

Answer 3

m = -3/2, so -m = 3/2

Answer 4

so y-(3) = 3/2 * (x-(-48)

Answer 5

y-3 = 3/2 * (x + 48)

Answer 6

distribute the 3/2, then add 3 to both sides

Answer 7

y = - 2/3 x - 29
correct

Answer 8

\[y = \frac{3}{2}x + (\frac{3}{2}\times 48) + 3\]

Answer 9

simplify

Answer 10

\[y = \frac{3}{2} + 72 + 3\]
\[y = \frac{3}{2} + 75\]

Answer 11

perpendicular to the original equation

Answer 12

going through that point (-48, 3)

Answer 13

x+75 you mean

Answer 14

yep

Answer 15

\[y=\frac{3}{2}x + 75\]

Answer 16

medal me

Answer 17

haa I have more questions but I just need you to correct me ...how do I medal you by the way

Answer 18

click best

Answer 19

and start a new question for each question

Answer 20

it seems like you are having trouble building your initial equation by rearranging the terms given

you need to isolate y

in this equation:
3x + 2y = -14

first you subtract 3x from both sides giving

2y = -3x -14

then you divide by 2, both sides

giving

y = (-3/2)x - 7

Answer 21

the coefficient on the x is your slope, m. in this case, it is -3/2. perpendicular lines have the slope with the opposite sign. so, perpendicular to this is 3/2

Answer 22

we took that slope value and plugged it into the point-slope equation:

y-y1 = m(x-x1)

giving y-y1 = (3/2)(x-x1)

where x1 and y1 and the x and y values from your point

Answer 23

ya get it?

Answer 24

now click "best answer"