Question

The line \((2+k)x+(1+k)y = 5+7k\) passes through the fixed point for different values of k. If 'd' is the distance of a fixed point from the origin, then \(d^2 = \ldots\)

• A. \(29\)
• B. \(37\)
• C. \(65\)
• D. \(85\)

Hint:

Group the terms by k to get two lines whose intersection is the fixed point.

Answer 1

The Correct Option is D

Step 1: Separate the parameter
Write \((2+k)x+(1+k)y = 5+7k\) as \((2x+y-5) + k(x+y-7) = 0\).

Step 2: Find the fixed point
For every \(k\) the line passes through the intersection of \(2x+y = 5\) and \(x+y = 7\). Subtracting the second from the first gives \(x = -2\), then \(y = 9\).

Step 3: Distance
The fixed point is \((-2,9)\). Then \(d^2 = (-2)^2 + 9^2 = 4+81 = 85\).

Step 4: Check
Test with the fixed point: \((2+k)(-2)+(1+k)(9) = -4-2k+9+9k = 5+7k\). It holds for all \(k\). So option (D).

Final Answer:
d squared equals 85. \[ \boxed{\text{(D)}\ 85} \]