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?
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?
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.
Find the slopes of the two lines and equate them to each other and solve for $p$.
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.
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 $y$.
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 $p=\frac{3}{5}$.