Animated Solution for Mathematics - Straight Lines: The lines $p(p^2+1)x-y+q=0$ and $(p^2+1{)}^{2}x+(p^2+1)y+2q=0$ are perpendicular to a common line for

We are presented with two lines, $L_1$ and $L_2$, defined by the equations:

The problem states that both lines are perpendicular to a common line, $L_3$. Geometrically, if two lines are perpendicular to the same line, they must be parallel to each other.

For $L_1$ and $L_2$ to be parallel, their slopes must be equal. Let the slopes of $L_1$ and $L_2$ be $m_1$ and $m_2$, respectively.

Using the standard form $Ax+By+C=0$, the slope is given by $m=-\frac{A}{B}$.

For $L_1$, we have $A=p(p^2+1)$ and $B=-1$. Therefore:

For $L_2$, we have $A=(p^2+1{)}^2$ and $B=(p^2+1)$. Therefore:

Since $L_1$ and $L_2$ are parallel, we set $m_1=m_2$:

Because $p$ is a real number, the term $(p^2+1)$ is always greater than or equal to $1$ and is never zero. We can safely divide both sides of the equation by $(p^2+1)$.

This simplification yields:

By leveraging the geometric property of parallel lines, we bypassed complex algebraic expansion. The only value that satisfies the given condition is $p=-1$.