# Thread: Find an equation of the line meeting the specified conditions

1. ## Find an equation of the line meeting the specified conditions

Containing the point (0,-3) and parallel to 4y = 8

Step 1: 4y div by 4y

Step 2: 8 div by 4

Step 3: y=2

Correct?

2. $4y = 8$
$y = 8/4$
$y = 2$

Y = 2 --> the y-value is constant, it'll always be the same.

If it's parallell, it'll be constant as well so...

the y- value of the point being -3

--> y = -3

3. hey there,

yup your answer is correct but i am not really sure about your step 1. I hope you meant divide by 4 instead of 4y. A better working is to use the general line equation:

$y = mx + c \longrightarrow$ given gradient and y-intercept

or,

$y-y_1 = m(x-x_1) \longrightarrow$ given a point the line passes and the gradient.

You are given a point (0,-3) and an equation of a line 4y = 8.

From the line equation given you can represent it in the general line equation of y = mx +c:

$
4y = 8
$

$
y = \frac 84 = 2
$

$
y = 0x + 2
$

gradient, m = 0 and y-intercept,c = 2

From this given line equation we can obtain the gradient of the required line equation since both lines are parallel.
Hence the gradient of the required line is 0.

Observing the information we have now it is more suitable to use the second formula to obtain the required line equation:

$y-y_1 = m(x-x_1)$
$y-(-3) = 0(x-0)$
$y = -3$