JEE Mathematics 2024 Question with Solution

Given: $a, b, c$ are three distinct consecutive terms of an A.P., and the lines $ax + by + c = 0$ are concurrent at a fixed point $P$.

Also, the system

has infinitely many solutions.

Find: The value of $(PQ)^2$.

Since $a, b, c$ are in A.P.,

so,

This shows that the line $ax + by + c = 0$ passes through the fixed point obtained by taking $x = 1$ and $y = -2$, because

Hence,

For the given system to have infinitely many solutions, the determinant of the coefficient matrix must be zero:

From the solution working,

Now use the determinant condition for the constants:

Substituting $\alpha = 8$, we get

Therefore,

Now compute the square of the distance between $P(1, -2)$ and $Q(8, 6)$:

Therefore, the value of $(PQ)^2$ is $113$.

Given: The lines $ax + by + c = 0$ correspond to three consecutive A.P. terms $a, b, c$, and the system has infinitely many solutions.

Find: $(PQ)^2$.

Because $a, b, c$ are in A.P.,

which gives

So the common point lies on every line $ax + by + c = 0$. Substituting $x = 1$ and $y = -2$ gives

Hence the fixed point is

For infinitely many solutions,

Expanding,

So,

Now use the determinant formed by replacing the first column with constants:

Expanding,

Thus,

and so

Finally,

Therefore, the required value is $113$.