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Find the value of k, so that the following lines intersect at the same point:

$$3x + y - 2 = 0$$ $$kx + 2y - 3 = 0$$ $$2x - y + 3 = 0$$

How can I resolve this? thanks

I was able to find that $(-\frac15,\frac{13}5)$ is the intersection of the first and the third line.

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closed as off-topic by Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh Dec 10 '18 at 12:03

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  • $\begingroup$ Welcome to MSE! What have you tried so far? $\endgroup$ – platty Dec 10 '18 at 5:40
  • $\begingroup$ I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct. $\endgroup$ – englishworkvgs Dec 10 '18 at 5:46
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    $\begingroup$ Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point? $\endgroup$ – platty Dec 10 '18 at 5:47
  • $\begingroup$ I got this result: -1/5, 13/5 But what should I do to get k and have the condition? $\endgroup$ – englishworkvgs Dec 10 '18 at 6:02
  • $\begingroup$ sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line. $\endgroup$ – 1ENİGMA1 Dec 10 '18 at 6:07
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Hint :

Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?

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  • $\begingroup$ Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well). $\endgroup$ – platty Dec 10 '18 at 8:06

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