Key Concepts and Formulas

• General Equation of a Line: The general equation of a line is given by $px + qy + r = 0$, where $p, q, r$ are constants and $x, y$ are variables.
• Concurrency of Lines: A set of lines is said to be concurrent if they all pass through the same point.
• Condition for Concurrency based on Coefficients: If a set of lines is given by $px + qy + r = 0$ and the coefficients satisfy a relation $Ap + Bq + Cr = 0$, then these lines are concurrent at the point $(-A/C, -B/C)$ if $C\neq0$. If $C=0$, the lines are parallel.

Step-by-Step Solution

Step 1: Understand the Given Information

We are given the general equation of a line: $px + qy + r = 0 \quad \text{(Equation 1)}$ and a condition that the coefficients $p, q, r$ must satisfy: $3p + 2q + 4r = 0 \quad \text{(Equation 2)}$ Our goal is to determine if the lines represented by Equation (1), subject to the constraint in Equation (2), are concurrent, parallel, or neither.

Step 2: Manipulate the Condition to Match the Line Equation Form

We want to express Equation (2) in a form that allows us to directly compare it with Equation (1). To do this, we isolate $r$ in Equation (2): $4r = -3p - 2q$ $r = -\frac{3}{4}p - \frac{2}{4}q$ $r = -\frac{3}{4}p - \frac{1}{2}q \quad \text{(Equation 3)}$

Step 3: Substitute the value of 'r' from Equation (3) into Equation (1)

Substitute the expression for $r$ from Equation (3) into Equation (1): $px + qy + \left(-\frac{3}{4}p - \frac{1}{2}q\right) = 0$ $px - \frac{3}{4}p + qy - \frac{1}{2}q = 0$ $p\left(x - \frac{3}{4}\right) + q\left(y - \frac{1}{2}\right) = 0 \quad \text{(Equation 4)}$

Step 4: Analyze Equation (4) to Determine Concurrency

Equation (4) must hold for all values of $p$ and $q$ that satisfy the original condition. This is only possible if: $x - \frac{3}{4} = 0$ and $y - \frac{1}{2} = 0$ Solving for $x$ and $y$, we get: $x = \frac{3}{4}$ $y = \frac{1}{2}$ This indicates that all lines pass through the point $\left(\frac{3}{4}, \frac{1}{2}\right)$.

Step 5: Conclude on Concurrency and Evaluate Options

Since all lines $px + qy + r = 0$ satisfying the condition $3p + 2q + 4r = 0$ invariably pass through the single fixed point $\left( \frac{3}{4}, \frac{1}{2} \right)$, the lines are concurrent at this point.

Let's examine the given options:

• (A) The lines are not concurrent: This is false, as we found a point of concurrency.
• (B) The lines are concurrent at the point $\left( {{3 \over 4},{1 \over 2}} \right)$: This matches our derived point of concurrency. This statement is true.
• (C) The lines are all parallel: For lines to be parallel, their slopes ($-p/q$) must be constant. The condition $3p + 2q + 4r = 0$ does not fix the ratio $p/q$. Thus, the lines are not all parallel.
• (D) Each line passes through the origin: This would mean the fixed point is $(0,0)$. However, our fixed point is $\left( \frac{3}{4}, \frac{1}{2} \right)$, which is not the origin. So, this statement is false.

Therefore, the correct statement is (B).

Common Mistakes & Tips

• Isolating r: While isolating 'r' works, it's more general to directly compare coefficients as shown in the Key Concepts.
• Checking for Parallelism: Remember that if the coefficient of r in the constraint equation is zero, the lines will be parallel, not concurrent.
• Sign Errors: Be extremely careful with signs when manipulating equations.

Summary

Given the equation of a line $px + qy + r = 0$ and a linear relation between its coefficients $3p + 2q + 4r = 0$, we found that the lines are concurrent. By manipulating the condition and comparing coefficients, we identified the point of concurrency as $\left(\frac{3}{4}, \frac{1}{2}\right)$. This analysis demonstrates that the lines are indeed concurrent at the specified point.

Final Answer The final answer is \boxed{The lines are concurrent at the point $\left( {{3 \over 4},{1 \over 2}} \right)$} which corresponds to option (B).