Question

If the three points $(3q,0),(0,3p),$ and $\left(1,1\right)$ are collinear, then which one is correct?

• A. $\frac{1}{p}+\frac{1}{q}=0$
• B. $\frac{1}{p}+\frac{1}{q}=1$
• C. $\frac{1}{p}+\frac{1}{q}=3$
• D. $\frac{1}{p}+\frac{3}{q}=1$

Answer 1

The correct option is C

$\frac{1}{p}+\frac{1}{q}=3$

Explanation for the correct option:

Step 1. Understanding the problem:

It is given that three points $A(3q,0),B(0,3p),$ and $C\left(1,1\right)$ are collinear.

Now as the three points are collinear so the slope of any two points must be the same. So, $m_{AB}=m_{BC}=m_{CA}$.

Step 2. Find the slope $m_{AB}$ and $m_{BC}$.

The slope of the line passing through $A(3q,0)$ and $B(0,3p)$ is given as:

$$\begin{gathered} m_{AB}=\frac{3p-0}{0-3q} \\ {m}_{AB}=\frac{3p}{-3q} \\ {m}_{AB}=-\frac{p}{q} \end{gathered}$$

The slope of the line passing through $B(0,3p)$ and $C(1,1)$ is given as:

$$\begin{gathered} m_{BC}=\frac{1-3p}{1-0} \\ {m}_{BC}=1-3p \end{gathered}$$

Step 3. Form and simplify the equation.

The two slopes must be the same, so:

$\begin{array}{rcl}{m}_{AB}& =& m_{BC}\\ -\frac{p}{q}& =& 1-3p\end{array}$

Take $-p$ common from both sides and cancel the terms.

$\begin{array}{rcl}-\frac{p}{q}& =& 1-3p\\ -p\left(\frac{1}{q}\right)& =& -p\left(-\frac{1}{p}+3\right)\\ \frac{1}{q}& =& -\frac{1}{p}+3\end{array}$

Now add $\frac{1}{p}$ both sides.

$$\begin{gathered} \begin{array}{l}\frac{1}{q}\end{array}+\frac{1}{p}=\begin{array}{l}-\frac{1}{p}\end{array}\begin{array}{l}+3\end{array}+\frac{1}{p} \\ \frac{1}{p}+\frac{1}{q}=3 \end{gathered}$$

Hence, the correct option is C.