JEE Mathematics 2025 Question with Solution

Given: The family of lines is $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$.

Find: The value of $d^2$, where $d$ is the distance between the common point $P$ and $Q(3, 6)$.

Rearrange the equation by grouping the terms containing $\lambda$:

A family of lines of the form $\lambda L_1 + L_2 = 0$ passes through the intersection of

So the common point $P$ is the intersection of

and

Multiply the second equation by $3$:

Subtracting from the first equation,

Hence,

Substitute $y = 2$ into $x + 2y - 5 = 0$:

Therefore,

Now use the distance formula between $P(1, 2)$ and $Q(3, 6)$:

Therefore,

So, the correct option is A.

Given: The line is $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$.

Find: The common point of the family and then compute $d^2$.

Let the common point be $P(h, k)$. Since it lies on the line for every value of $\lambda$, substitute $x = h$ and $y = k$:

For this to be true for all $\lambda$, both coefficients must be zero:

Solve these equations:

Multiply the second equation by $3$:

Subtracting,

Then,

So,

Now calculate the distance from $P(1, 2)$ to $Q(3, 6)$:

Hence,

Therefore, the correct option is A.

The second extracted approach on the solution's computes the distance from $Q$ to the line $3x + 7y = 17$ and obtains the same numerical value $20$ for $d^2$. The accepted answer from the source is A, and the standard solution above directly follows the stated distance between points $P$ and $Q$.