Question

1. the graphs of $y = px + q$ and $y = rx + s$ are perpendicular and $p = 4|r|$. find the values of $p$ and $r$.

Explanation:

Step1: Recall the slope condition for perpendicular lines

For two lines with slopes \(m_1\) and \(m_2\) to be perpendicular, the product of their slopes is \(-1\), i.e., \(m_1\times m_2=-1\). Here, the slopes of the lines \(y = px + q\) and \(y=rx + s\) are \(p\) and \(r\) respectively. So we have the equation \(p\times r=-1\).

Step2: Use the given relation \(p = 4|r|\)

We know that \(p = 4|r|\). Let's consider two cases for the absolute value:

• Case 1: If \(r\geq0\), then \(|r| = r\), so \(p = 4r\). Substitute into \(p\times r=-1\), we get \(4r\times r=-1\), i.e., \(4r^{2}=-1\). But the square of a real number is non - negative, and \(4r^{2}\geq0\), while \(- 1<0\), so this case has no real solutions.
• Case 2: If \(r<0\), then \(|r|=-r\), so \(p = 4(-r)=-4r\). Substitute into \(p\times r=-1\), we have \((-4r)\times r=-1\), which simplifies to \(-4r^{2}=-1\). Divide both sides by \(- 4\): \(r^{2}=\frac{1}{4}\). Take the square root of both sides, since \(r < 0\), \(r=-\frac{1}{2}\).

Step3: Find the value of \(p\)

Now that we know \(r = -\frac{1}{2}\), substitute into \(p=-4r\) (from the case when \(r < 0\)). Then \(p=-4\times(-\frac{1}{2}) = 2\). We can also check the product \(p\times r=2\times(-\frac{1}{2})=-1\), which satisfies the perpendicular slope condition.

Answer:

\(p = 2\) and \(r=-\frac{1}{2}\)