Question

If a straight line drawn through the point of intersection of the lines \(4x + 3y - 1 = 0\) and \(3x + 4y - 1 = 0\), meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:

• A. \(x + y - 7 = 0\)
• B. \(x + y - 14xy = 0\)
• C. \(2x + y + 14xy = 0\)
• D. \(x + 2y - 14xy = 0\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We must first find the point of intersection of the two given lines. Then, a generic line passing through this point intercepts the coordinate axes at points P and Q. The goal is to find the mathematical relationship (locus) defined by the coordinates of the midpoint of the line segment PQ.
Step 2: Key Formula or Approach:
1. Solve the system of linear equations to find the intersection point \( (x_0, y_0) \).
2. Assume the equation of the line in intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \).
3. The axis intersections are \( P(a, 0) \) and \( Q(0, b) \).
4. The midpoint of PQ is \( (h, k) = (\frac{a}{2}, \frac{b}{2}) \).
5. Substitute \( a \) and \( b \) in terms of \( h \) and \( k \) into the line equation satisfying the intersection point to obtain the locus.
Step 3: Detailed Explanation:
Let's find the point of intersection of:
\( 4x + 3y = 1 \) \quad \dots(1)
\( 3x + 4y = 1 \) \quad \dots(2)
Adding (1) and (2):
\( 7x + 7y = 2 \implies x + y = \frac{2}{7} \).
Subtracting (2) from (1):
\( x - y = 0 \implies x = y \).
Substitute \( x = y \) into the addition result:
\( x + x = \frac{2}{7} \implies 2x = \frac{2}{7} \implies x = \frac{1}{7} \) and thus \( y = \frac{1}{7} \).
The point of intersection is \( (\frac{1}{7}, \frac{1}{7}) \).
Let the straight line have x-intercept \( a \) and y-intercept \( b \). Its equation is:
\( \frac{x}{a} + \frac{y}{b} = 1 \).
Since the line passes through \( (\frac{1}{7}, \frac{1}{7}) \), substitute these coordinates:
\( \frac{1/7}{a} + \frac{1/7}{b} = 1 \implies \frac{1}{a} + \frac{1}{b} = 7 \).
The line intersects the axes at \( P(a, 0) \) and \( Q(0, b) \).
Let the midpoint of PQ be \( (h, k) \).
By the midpoint formula: \( h = \frac{a + 0}{2} = \frac{a}{2} \implies a = 2h \).
And \( k = \frac{0 + b}{2} = \frac{b}{2} \implies b = 2k \).
Substitute \( a \) and \( b \) into our equation:
\( \frac{1}{2h} + \frac{1}{2k} = 7 \).
Multiply the entire equation by 2:
\( \frac{1}{h} + \frac{1}{k} = 14 \).
Find the common denominator:
\( \frac{h + k}{hk} = 14 \implies h + k = 14hk \).
To represent the locus, replace \( (h, k) \) with generic coordinates \( (x, y) \):
\( x + y = 14xy \implies x + y - 14xy = 0 \).
Step 4: Final Answer:
The locus of the midpoint is \( x + y - 14xy = 0 \).