Math CentralQuandaries & Queries


Hi, Im in yr 11 at school and Im having trouble with a maths question on one of my assignments. The question is as follows:

"The line with the equation 3x + by = 6 intersects with the line 6y + ax = c at right angles at the point (4,6). Determine the values of a, b & c."

I've found the value of b (-1, I think) however i have NO idea how to find a or c. if you could help me
that would be very much appreciated. Thanks very much

Hi Ellen.

You're right: b is -1. You get this by noting that the first line must pass through (4, 6) so substitute 4 and 6 in for x and y in the first equation to solve for b:

3(4) + b(6) = 6
6b = 6 - 12
b = -1

Two lines on a graph that meet at right angles are perpendicular to each other, so their slopes are the negative reciprocal of one another. For example, if the slope of one line is m, then the slope of the other is -1/m.

Let's re-write the lines in y=mx+b form to reveal the slopes:

3x - y = 6 becomes y = 3x - 6 and
6y + ax = c becomes y = (-a/6)x + c/6.

So if these slopes are the negative reciprocal of one another,

(-a/6) = -1 / (3) = -1/3.

Now you can solve for a. And with the values for a, x, and y, you can easily find the remaining unknown c by substituting these values into the equation for the second line.

Hope this helps,
Stephen La Rocque.

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