Math CentralQuandaries & Queries


Question from James, a student:

I need to put it in Standard from ax+by=c the problem is
(-4,0); parallel to y=-2+1 how can I put this in Standard from

Hi James. Let me illustrate with a similar question.

Q. Write the standard form of the equation of the line that passes through (-2, 3) and is perpendicular to
y = (1/3)x - 4.

A. I know that two lines that are perpendicular have slopes that are the negative reciprocal of each other. In your problem, you have parallel lines (they have the same slope). So first I look at the slope of the given line: y = (1/3)x - 4 is already in "slope-intercept" form, so I can see that the slope is (1/3) because the slope is the m when your equation is in the y = mx+b form.

Therefore the slope of a line that is perpendicular to this one is -3/1 (the negative reciprocal). If the question said "parallel", then I'd just say the slope of the new line is the same as the other one.

The slope of a line in standard form Ax + By = C is -A/B. So in my case, -3/1 = -A/B, so I can take A = 3 and B = 1 and I have

3x + 1y = C

To find the value of C, I substitute in the point I am given for (x, y). That is,

3(-2) + 1(3) = C
-6 + 3 = C
C = -3.

Now I can write the answer: 3x + y = -3.

Try the same approach with your question James.

Stephen La Rocque.

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