Question

Every point on the curve \(3x + 2y - 3xy = 0\) is the centroid of a triangle formed by the coordinate axes and a line (L) intersecting both the coordinate axes. Then all such lines (L)

• A. are parallel
• B. are concurrent
• C. intersect each other at different points
• D. are perpendicular to the tangents to the curve

Hint:

If the parameters \(a, b\) of a line \(ax + by + c = 0\) (or intercept form) satisfy a linear relation \(Aa + Bb + C = 0\), the line passes through a fixed point.

Answer 1

The Correct Option is B

Step 1: Setup the Centroid condition:
Let the line L have intercepts \(a\) and \(b\). Its equation is \(\frac{x}{a} + \frac{y}{b} = 1\). The vertices of the triangle are \((0,0), (a,0), (0,b)\). The centroid \(G(\bar{x}, \bar{y})\) is given by: \[ \bar{x} = \frac{0+a+0}{3} = \frac{a}{3} \implies a = 3\bar{x} \] \[ \bar{y} = \frac{0+0+b}{3} = \frac{b}{3} \implies b = 3\bar{y} \]
Step 2: Use the Locus Equation:
The centroid \((\bar{x}, \bar{y})\) lies on the curve \(3x + 2y - 3xy = 0\). \[ 3\bar{x} + 2\bar{y} - 3\bar{x}\bar{y} = 0 \] Divide by \(\bar{x}\bar{y}\) (assuming \(\bar{x}, \bar{y} \neq 0\)): \[ \frac{3}{\bar{y}} + \frac{2}{\bar{x}} = 3 \]
Step 3: Relate to Line Parameters:
Substitute \(\bar{x} = a/3\) and \(\bar{y} = b/3\): \[ \frac{3}{b/3} + \frac{2}{a/3} = 3 \] \[ \frac{9}{b} + \frac{6}{a} = 3 \] Divide by 3: \[ \frac{3}{b} + \frac{2}{a} = 1 \quad \text{or} \quad \frac{2}{a} + \frac{3}{b} = 1 \]
Step 4: Analyze Line Family:
The equation of line L is \(\frac{x}{a} + \frac{y}{b} = 1\). The condition derived is \(\frac{2}{a} + \frac{3}{b} = 1\). This implies that the line always satisfies the equation for the point \(x=2, y=3\). Thus, all such lines pass through the fixed point \((2, 3)\). Hence, they are concurrent.