Write an equation of the line that is parallel to the given line and that passes through the given point. 3x + 2y = 5; (2, 6)

Answer 1

The first step I would do toward solving this problem is to get the equation in slope-intercept form. This means isolating y as a function of x.

3x + 2y = 5

2y = -3x + 5

The next step involves dividing every term by 2 to render y alone:

y = (-3/2)x + 5/2

This is the equation of the line to which we want to find a parallel line. Remember that the slope of a line parallel another is the same slope. In slope intercept form, the slope of the line is the coefficient of the x, so -3/2 in this problem.

So now we must find the y-intercept.

I usually think of this in two ways. Firstly, you can sketch a graph and plot a point at (2,6), and since you know the slope (rise/run) is -3/2, you can work backwards to find the y-intercept.

The algebraic way to do this is to put the point and slope into point-slope form: y-y₁ = m(x-x₁)

where m is the slope and (x₁,y₁) is your point. Plugging this in you get:

y - 6 = -3/2(x - 2)

Now you only have to solve for y in slope-intercept form

y - 6 = (-3/2)x + 6/2

y - 6 = (-3/2)x + 3

y = (-3/2)x + 9

This is your final answer, the equation of a line parallel to 3x + 2y = 5 and passing through the point (2,-6) in slope-intercept form is

y = (-3/2)x + 9