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: Intersection equation
Substituting $y = 3 - x$ gives $5x^2 - 15x + 9 = 0$.

Step 2: Sum of roots
$x_1 + x_2 = 15/5 = 3$.

Step 3: Midpoint
$x_m = 3/2$ and $y_m = 3 - 3/2 = 3/2$. Option (D).

Final Answer:
Option (D). \[ \boxed{\left(\frac{3}{2},\frac{3}{2}\right)} \]