So I'm supposed to find the value of m for which these three lines intersect at the same point:

\(\displaystyle mx+2y-1=0\)

\(\displaystyle 2x+my+3=0\)

\(\displaystyle x-y-3=0\)

One of the many things I tried was to write all three in slope-intercept form like this:

\(\displaystyle y=(1-mx)/2\)

\(\displaystyle y=(-3-2x)/m\)

\(\displaystyle y=x-3\)

Then I thought, I can just equate the second and third equation since I know for sure that the lines do intersect, and that left me with this mess:

\(\displaystyle (-3-2x)/m=x-3\)

\(\displaystyle -3-2x=mx-3\)

\(\displaystyle mx+2x=0\)

\(\displaystyle x(m+2)=0\)

\(\displaystyle x=0\text{ and }m = -2\)

Now I'm pretty damn sure that this is not the solution, in fact I'm convinced that my whole method of going into this is wrong.

\(\displaystyle mx+2y-1=0\)

\(\displaystyle 2x+my+3=0\)

\(\displaystyle x-y-3=0\)

One of the many things I tried was to write all three in slope-intercept form like this:

\(\displaystyle y=(1-mx)/2\)

\(\displaystyle y=(-3-2x)/m\)

\(\displaystyle y=x-3\)

Then I thought, I can just equate the second and third equation since I know for sure that the lines do intersect, and that left me with this mess:

\(\displaystyle (-3-2x)/m=x-3\)

\(\displaystyle -3-2x=mx-3\)

\(\displaystyle mx+2x=0\)

\(\displaystyle x(m+2)=0\)

\(\displaystyle x=0\text{ and }m = -2\)

Now I'm pretty damn sure that this is not the solution, in fact I'm convinced that my whole method of going into this is wrong.

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