Question

find the equation of a line that is perpendicular to 2x-3y=3 and contains the points (0,-3), write your answer ina slope intercept form

Answer 1

A perpendicular equation must have what in different?

Answer 2

i don't understand ur ?

Answer 3

A perpendicular line has a negative reciprocal and different y-interecepts to the given equation.

Answer 4

Lets turn your equation into a y = mx + b format
2x - 3y = 3
-3y = -2x + 3
y = 2/3x - 1

Answer 5

y=-3x/2-3

Answer 6

What is the equation of the given line, britt?

Answer 7

this the whole question?

Answer 8

Your slope therefore is 2/3.

In your new equation, your slope is going to be -3/2 as it is perpendicular, and perpendicular lines have negative reciprocal slopes from your original equation

Answer 9

find the equation of a line that is perpendicular to 2x-3y=3 and contains the points (0,-3), write your answer ina slope intercept form

Okay-

First, we want to rearrange the equation. 2x - 3y = 3 | We're rearranging to get 'y' to be alone, so in the form y=mx+c

So, if you rearrange, you get y = (2x/3) - 1

Now, you want to find a purpendicular line that goes through 0,-3.

This means that the gradient must add up to -1. So, we're going to use the gradient -3/2, as it adds up to -1.

So, our new form of equation is y = -3x/2 + c

We want to find out what c is, as we don't know. So we put in what we have, 0 as x, and -3 as y.

So -3 = c

Now, the final equation is; y = -3x/2 - 3

Or, in another form, 2x + 3x + 6 = 0

Answer 10

Lets plug in the slope we found into our new equation:
y = mx + b
y = (-3/2) x + b

We also have your given points which are (0, - 3) = (x1,y1)

Now plug all your information in this equation:
(y -y1) = m (x-x1)
y + 3 = -3/2 ( x- 0)
y + 3 = -3/2x + 0
y = -3/2x + 3

Answer 11

y=-3x/2-3