Question

Write the equation of the line that passes through the points (3,2) and is parallel to the line defined by 3y+x=6 can someone answer this step by step

Answer 1

y = -⅓x + 3

Explanation:

By definition, parallel lines have the same slope.

Given the linear equation, 3y + x = 6, and the point, (3, 2):

Transform the given equation into its slope-intercept form, y = mx +b, where m = slope, and b = y-intercept.

3y + x = 6

Subtract x from both sides:

3y + x - x = - x + 6

3y = -x + 6

Divide both sides by 3 to isolate y:

`(3y)/(3) = (- x + 6)/(3)`

y = - ⅓x + 2 (slope, m = - ⅓, and y-intercept, b = 2).

Since the slope of the given line is - ⅓, then it means that the other line must also have the same slope of - ⅓. All we need to do at this point is to determine the y-intercept of the other line. Using the slope, m = -⅓, and the given point, (3, 2), substitute their values into the slope-intercept form:

y = mx + b

2 = -⅓(3) + b

2 = -1 + b

Add 1 to both sides to isolate b:

2 + 1 = -1 + 1 + b

3 = 0 + b

3 = b

Therefore, the equation of the other line that is parallel to 3y + x = 6 is:

y = -⅓x + 3.