Question

The line \(x+y = 3\) intersects the pair of straight lines \(x^2-3xy+y^2 = 0\) at points A and B. Then the co-ordinates of the mid-point of AB are

• A. \((\frac{5}{2},\frac{3}{2})\)
• B. \((\frac{1}{2},\frac{1}{2})\)
• C. \((-\frac{3}{2},\frac{1}{2})\)
• D. \((\frac{3}{2},\frac{3}{2})\)

Hint:

Substitute y = 3 - x into the pair of lines and use the sum of the roots.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The pair of lines \(x^2 - 3xy + y^2 = 0\) meets the line \(x + y = 3\) at two points A and B. The midpoint x coordinate is half the sum of the roots of the resulting quadratic.

Step 2: Substitute
With \(y = 3 - x\):
\[ x^2 - 3x(3 - x) + (3 - x)^2 = 0 \]
\[ x^2 - 9x + 3x^2 + 9 - 6x + x^2 = 0 \Rightarrow 5x^2 - 15x + 9 = 0 \]

Step 3: Midpoint
Sum of roots \(x_1 + x_2 = 3\), so midpoint x is \(\frac32\). From the line, \(y = 3 - \frac32 = \frac32\).
The midpoint is \(\left(\frac32, \frac32\right)\). It lies on \(x + y = 3\); option (A), (5/2, 3/2), has sum 4 and (B) has sum 1, so they are not on the line.

Final Answer:
The midpoint of AB is \(\left(\frac32, \frac32\right)\), option (D). \[ \boxed{\left(\frac{3}{2},\frac{3}{2}\right)} \]