Find the x- and y-intercepts for the line given by the equation 3x + 2y = 12 Find the equation of the line that passes through the points (3,1) and (2,-1).

Question

These are math problems I have challenge in knowing how to know which principles are applicable and how to apply them properly and how to write the equation in slope-intercept form. The answer to my earlier post had save me a whole lot of doubts, I appreciate. 

Answer 1

3x + 2y = 12

 

The x-intercept is the point where the line crosses the x-axis and occurs when y=0, so:

 

3x + 2(0) = 12             [Plugged y=0 into the equation 3x + 2y = 12]

3x = 12                      

x = 4                           [Divided both sides by 3]

 

So the x-intercept occurs at the point (x=4,y=0), or just (4,0)

 

The y-intercept is the point where the line crosses the y-axis and occurs when x=0, so:

 

3(0) + 2y = 12              [Plugged x=0 into the equation 3x + 2y = 12]

2y = 12

y = 6                            [Divided both sides by 2]

 

So the y-intercept occurs at the point (0,6)

 

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To find the equation of the line that passes through (3,1) and (2,-1), use the slope-intercept form for a straight line:

 

y = mx + b

 

Where m is the slope of the line and b is its y-intercept.  So we need to find the values of m and b to get the equation for this particular line.  The slope, m, is defined as the change in y over the change in x between any two points, (x₁,y₁) and (x₂,y₂), on the line:

 

m = (y₂-y₁) / (x₂-x₁)

 

In your case, you have two points, (2,-1) and (3,1).  So:

 

m = (1-(-1)) / (3-2) = 2/1 = 2

 

So far, then, our equation is:

 

y = 2x + b

 

To find b, plug the values of either point into y=2x+b and solve for b.  I'll use (3,1):

 

1 = 2(3) + b        [Plugged x=3 and y=1 into y = 2x + b)

1 = 6 + b

 

-5 = b                [Subtracted 6 from both sides of the equation]

 

So our equation for the line is y = 2x - 5