Question

$L_1: 3x + y = 0$

$L_2: 4x + 3y + 5 = 0$

Consider family of straight lines.

$L_1 + \lambda L_2 = 0, \lambda \in R$

Match List-I with List-II and select the correct answer using the code given below the list

Answer 1

P → 3, Q → 4, R → 1, S → 2

Answer 2

The problem involves a family of straight lines $L_1 + \lambda L_2 = 0$, where $L_1: 3x + y = 0$ and $L_2: 4x + 3y + 5 = 0$.
The family of lines passes through the point of intersection of $L_1$ and $L_2$.
To find the point of intersection:
From $L_1$, $y = -3x$.
Substitute into $L_2$: $4x + 3(-3x) + 5 = 0 \Rightarrow 4x - 9x + 5 = 0 \Rightarrow -5x + 5 = 0 \Rightarrow x = 1$.
Then $y = -3(1) = -3$.
So, the fixed point through which all lines in the family pass is $P(1, -3)$.

Let's evaluate each item:

(P) thrice of absolute value of y-intercept of line having slope $-\frac{[\pi+1]}{[e+1]}$ (where [.] represent greatest integer function)

First, calculate the value of the slope:
$[\pi+1] = [3.14159... + 1] = [4.14159...] = 4$.
$[e+1] = [2.71828... + 1] = [3.71828...] = 3$.
The given slope is $m = -\frac{4}{3}$.
The family of lines is $(3x + y) + \lambda(4x + 3y + 5) = 0$, which can be written as $(3+4\lambda)x + (1+3\lambda)y + 5\lambda = 0$.
The slope of a line in this family is $m = -\frac{3+4\lambda}{1+3\lambda}$.
If we set this equal to $-\frac{4}{3}$:
$-\frac{3+4\lambda}{1+3\lambda} = -\frac{4}{3} \Rightarrow 3(3+4\lambda) = 4(1+3\lambda) \Rightarrow 9+12\lambda = 4+12\lambda \Rightarrow 9 = 4$.
This is a contradiction, meaning no line of the form $L_1 + \lambda L_2 = 0$ for a finite $\lambda$ has this slope. However, the family of lines $L_1 + \lambda L_2 = 0$ represents all lines passing through the intersection point except for $L_2=0$. The line $L_2: 4x + 3y + 5 = 0$ itself has a slope of $-\frac{4}{3}$. Therefore, $L_2$ is the line in question.
For $L_2: 4x + 3y + 5 = 0$, to find the y-intercept, set $x=0$:
$3y + 5 = 0 \Rightarrow y = -5/3$.
Thrice of the absolute value of the y-intercept is $3 \times |-5/3| = 3 \times (5/3) = 5$.
So, (P) matches with (3).

(Q) Square of maximum distance of point (2, -6) from any member of family is

The family of lines passes through the fixed point $P(1, -3)$. Let the given point be $A(2, -6)$.
The distance from a point $A$ to a line passing through a fixed point $P$ is maximized when the line is perpendicular to the segment $AP$. In this case, the maximum distance is simply the length of the segment $AP$.
Distance $AP = \sqrt{(2-1)^2 + (-6 - (-3))^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.
The square of the maximum distance is $(\sqrt{10})^2 = 10$.
So, (Q) matches with (4).

(R) If locus of foot of perpendicular drawn from (0, 0) to any member of family is $x^2 + y^2 + ax + by$, then (a + b) is

The family of lines passes through the fixed point $P(1, -3)$. The foot of the perpendicular is drawn from the origin $O(0, 0)$.
The locus of the foot of the perpendicular from a fixed point $(x_0, y_0)$ to a family of lines passing through another fixed point $(x_1, y_1)$ is a circle with the segment connecting $(x_0, y_0)$ and $(x_1, y_1)$ as its diameter.
Here, $(x_0, y_0) = (0, 0)$ and $(x_1, y_1) = (1, -3)$.
The equation of the circle is $(x - x_0)(x - x_1) + (y - y_0)(y - y_1) = 0$.
$(x - 0)(x - 1) + (y - 0)(y - (-3)) = 0$
$x(x - 1) + y(y + 3) = 0$
$x^2 - x + y^2 + 3y = 0$
$x^2 + y^2 - x + 3y = 0$.
Comparing this with $x^2 + y^2 + ax + by = 0$, we get $a = -1$ and $b = 3$.
Then $a + b = -1 + 3 = 2$.
So, (R) matches with (1).

(S) If $(\alpha, \beta)$ be the image of (0, 0) with respect to $L_2$, then $(\alpha^2 + \beta^2)$ is

The line $L_2$ is $4x + 3y + 5 = 0$. The point is $(x_1, y_1) = (0, 0)$.
The formula for the image $(\alpha, \beta)$ of a point $(x_1, y_1)$ with respect to the line $Ax + By + C = 0$ is:
$\frac{\alpha - x_1}{A} = \frac{\beta - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2}$.
Here, $A=4, B=3, C=5$, and $(x_1, y_1) = (0, 0)$.
$\frac{\alpha - 0}{4} = \frac{\beta - 0}{3} = -2 \frac{4(0) + 3(0) + 5}{4^2 + 3^2}$
$\frac{\alpha}{4} = \frac{\beta}{3} = -2 \frac{5}{16 + 9}$
$\frac{\alpha}{4} = \frac{\beta}{3} = -2 \frac{5}{25}$
$\frac{\alpha}{4} = \frac{\beta}{3} = -\frac{10}{25} = -\frac{2}{5}$.
From this, $\alpha = 4 \times (-\frac{2}{5}) = -\frac{8}{5}$.
And $\beta = 3 \times (-\frac{2}{5}) = -\frac{6}{5}$.
We need to find $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = \left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = \frac{64}{25} + \frac{36}{25} = \frac{64 + 36}{25} = \frac{100}{25} = 4$.
So, (S) matches with (2).

Final Matching: (P) - (3) (Q) - (4) (R) - (1) (S) - (2)