Question

If $px+qy+r=0$ represent a family of straight lines such that $3p+2q+4r=0$ then

• A. All lines are parallel
• B. All lines are inconsistent
• C. All lines are concurrent at $\left(\frac{3}{4},\frac{1}{2}\right)$
• D. All lines are concurrent at $(3,2)$

Answer 1

The correct option is C

All lines are concurrent at $\left(\frac{3}{4},\frac{1}{2}\right)$

Explanation for the correct option.

Given that, $px+qy+r=0....\left(1\right)$ is a family of straight lines and $3p+2q+4r=0$.

Using the given two conditions we have :

$\frac{3p}{4}+\frac{q}{2}+r=0...\left(2\right)$

$\begin{array}{l}\Rightarrow \frac{3p}{4}+\frac{2}{4}q+\frac{4r}{4}=0\\ \Rightarrow \frac{3}{4}p+\frac{1}{2}q+r=0\end{array}$

Now, compare the above equation with the set of all line equations,

$\begin{array}{l}p\left(\frac{3}{4}\right)+q\left(\frac{1}{2}\right)+r=0\\ \Rightarrow x=\frac{3}{4}\text{and}y=\frac{1}{2}\end{array}$

Now using the condition of concurrency of lines (1) & (2) we have:

$\frac{x}{{\displaystyle \frac{3}{4}}}=\frac{y}{{\displaystyle \frac{1}{2}}}=1$

The point $\left(\frac{3}{4},\frac{1}{2}\right)$ satisfies the equation $px+qy+r=0$.

Thus, we can say that lines are concurrent at $\left(\frac{3}{4},\frac{1}{2}\right)$.

Hence option C is correct.