Let $P$ be any point on the line $x - y + 3 = 0$ and $A$ be a fixed point $(3, 4)$. If the family of lines given by the equations $(3\sec\theta + 5\cosec\theta)x + (7\sec\theta - 3\cosec\theta)y + 11(\sec\theta - \cosec\theta) = 0$ are concurrent at a point $B$ for all permissible value of $\theta$, then:
sum of the abscissa and ordinate of point $B$ is equal to $-1$.
product of the abscissa and ordinate of point $B$ is equal to $-2$.
maximum value of $|PA - PB|$ is $2\sqrt{10}$.
minimum value of $PA + PB$ is $2\sqrt{34}$.
Step-by-Step Solution
Key Concept: First, identify the point of concurrency for the family of lines by grouping terms involving common parameters (like secθ, cosecθ) and setting their coefficients to zero. Then, apply the triangle inequality for the maximum value of $|PA-PB|$ and the reflection principle for the minimum value of $PA+PB$ involving a point $P$ on a given line.
Step 1: The family of lines is given by the equation $(3\sec\theta + 5\cosec\theta)x + (7\sec\theta - 3\cosec\theta)y + 11(\sec\theta - \cosec\theta) = 0$.
To find the point of concurrency $B$, rearrange the equation by grouping terms involving $\sec\theta$ and $\cosec\theta$:
$$ \sec\theta(3x + 7y + 11) + \cosec\theta(5x - 3y - 11) = 0 $$
Step 2: For this equation to hold for all permissible values of $\theta$, the coefficients of $\sec\theta$ and $\cosec\theta$ must both be zero. This yields a system of two linear equations:
$$ 3x + 7y + 11 = 0 \quad (1) $$
$$ 5x - 3y - 11 = 0 \quad (2) $$
Step 3: Solve the system of equations for $x$ and $y$.
From (1), $3x + 7y = -11$.
From (2), $5x - 3y = 11$.
Multiply equation (1) by 3 and equation (2) by 7:
$$ 9x + 21y = -33 $$
$$ 35x - 21y = 77 $$
Adding these two new equations:
$$ (9x + 35x) + (21y - 21y) = -33 + 77 $$
$$ 44x = 44 $$
$$ x = 1 $$
Substitute $x=1$ into equation (1):
$$ 3(1) + 7y + 11 = 0 $$
$$ 3 + 7y + 11 = 0 $$
$$ 7y + 14 = 0 $$
$$ 7y = -14 $$
$$ y = -2 $$
Thus, the point of concurrency is $B = (1, -2)$.
Step 4: The sum of the abscissa and ordinate of point $B$ is $1 + (-2) = -1$.
Step 5: The product of the abscissa and ordinate of point $B$ is $1 \times (-2) = -2$.
Step 6: Calculate the distance $AB$ between $A(3, 4)$ and $B(1, -2)$.
$$ AB = \sqrt{(3-1)^2 + (4-(-2))^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} $$
Step 7: By the triangle inequality, for any point $P$, $|PA - PB| \leq AB$. The maximum value of $|PA - PB|$ is $AB = 2\sqrt{10}$.
Step 8: Determine the minimum value of $PA + PB$ for $P$ on the line $L: x - y + 3 = 0$.
First, check if points $A(3,4)$ and $B(1,-2)$ are on the same or opposite sides of the line $L$.
For point $A$: $3 - 4 + 3 = 2$.
For point $B$: $1 - (-2) + 3 = 1 + 2 + 3 = 6$.
Since both values are positive, $A$ and $B$ are on the same side of the line $L$.
To find the minimum value of $PA+PB$, reflect one point (say $B$) across the line $L$ to get $B'$. The minimum distance will be $AB'$.
Let $B=(1,-2)$ and the line be $x-y+3=0$. The coordinates of the reflected point $B'=(x', y')$ are given by:
$$ \frac{x' - 1}{1} = \frac{y' - (-2)}{-1} = -2 \frac{1(1) + (-1)(-2) + 3}{1^2 + (-1)^2} $$
$$ \frac{x' - 1}{1} = \frac{y' + 2}{-1} = -2 \frac{1 + 2 + 3}{1 + 1} $$
$$ \frac{x' - 1}{1} = \frac{y' + 2}{-1} = -2 \frac{6}{2} $$
$$ \frac{x' - 1}{1} = \frac{y' + 2}{-1} = -6 $$
From these equations:
$x' - 1 = -6 \Rightarrow x' = -5$.
$y' + 2 = 6 \Rightarrow y' = 4$.
So, the reflected point is $B' = (-5, 4)$.
The minimum value of $PA+PB$ is the distance $AB'$:
$$ AB' = \sqrt{(3 - (-5))^2 + (4 - 4)^2} = \sqrt{(3+5)^2 + 0^2} = \sqrt{8^2} = 8 $$
Correct Answer: 1, 2, 3, 4