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The lines

$px + (2p-1)y + 4 = 0$


$(p+3)x + 2py + 6 = 0$

are parallel to each other. Find $p$.

I have no idea how to tackle this problem, can anyone help?

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up vote 0 down vote accepted

As the two lines are parallel to each other, so the coefficients should satisfy that $\frac{p}{p+3}=\frac{2p-1}{2p}$, or $2p^{2}=(p+3)(2p-1)$, so $p=\frac{3}{5}$.

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Don't understand – JohnPhteven Oct 4 '12 at 16:49
@Zafars: It's a criterion for two lines parallel to each other or not. – Alfred Chern Oct 4 '12 at 16:54
Thank you, kind sir, I now understand. – JohnPhteven Oct 7 '12 at 7:48

Find the slopes of the two lines and equate them to each other and solve for $p$.

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How can I flind the slope of the line? – JohnPhteven Oct 4 '12 at 16:40
@ZafarS: If you don't already know that, then apply some independent thought. Can you find the axis intercepts with the equation of a line in this form? Can you find the slope of the line that connects those two intercept points? – Henning Makholm Oct 4 '12 at 21:43

If two lines are parallel then their slopes are equal.Equate slopes of two lines and you get a quadratic equation.Solve it for your answer.

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But, the important thing for this problem is "If two lines have the same slope, then they are parallel." Your statement gives the other half which is not important in this problem. – Graphth Oct 4 '12 at 16:37
I don't know how to get the slope of these lines, I am only familiar to the simple $ y =ax+b $ – JohnPhteven Oct 4 '12 at 16:41

Two lines in 2-dimensions are parallel if and only if they have the same slope. So, just find the slope of both lines. If you write both in the form $y = mx + b$, then $m$ is the slope. So, for both, solve for $y$.

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How do I do this? Can you give an example? I don't know how to write these in the form y=ax+b.. – JohnPhteven Oct 4 '12 at 16:38

Fact that lines $px + (2p-1)y + 4 = 0$ and $(p+3)x + 2py + 6 = 0$ are parallel one with other means that corresponding system of two above equations has no solutions, then apply Cramer's rule.

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I don't believe that we've reached that level of mathematics yet.. – JohnPhteven Oct 4 '12 at 16:49

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