The lines
$px + (2p-1)y + 4 = 0$
and
$(p+3)x + 2py + 6 = 0$
are parallel to each other. Find $p$.
I have no idea how to tackle this problem, can anyone help?
I have no idea how to tackle this problem, can anyone help? |
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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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Find the slopes of the two lines and equate them to each other and solve for $p$. |
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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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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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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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