Question:

Step 1: Determine the distance between the lines $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$. Let $u = x+7y$. The equation becomes $u^2 + 4\sqrt{2}u - 42 = 0$. Solving for $u$ using the quadratic formula: $$u = \frac{-4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4(1)(-42)}}{2(1)}$$ $$u = \frac{-4\sqrt{2} \pm \sqrt{32 + 168}}{2}$$ $$u = \frac{-4\sqrt{2} \pm \sqrt{200}}{2}$$ $$u = \frac{-4\sqrt{2} \pm 10\sqrt{2}}{2}$$ This yields two values for $u$: $$u_1 = \frac{-4\sqrt{2} + 10\sqrt{2}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$ $$u_2 = \frac{-4\sqrt{2} - 10\sqrt{2}}{2} = \frac{-14\sqrt{2}}{2} = -7\sqrt{2}$$ Thus, the two parallel lines are $x+7y - 3\sqrt{2} = 0$ and $x+7y + 7\sqrt{2} = 0$. The distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$. Here, $A=1$, $B=7$, $C_1=-3\sqrt{2}$, and $C_2=7\sqrt{2}$. $$D = \frac{|-3\sqrt{2} - (7\sqrt{2})|}{\sqrt{1^2+7^2}} = \frac{|-10\sqrt{2}|}{\sqrt{1+49}} = \frac{10\sqrt{2}}{\sqrt{50}} = \frac{10\sqrt{2}}{5\sqrt{2}} = 2$$ Step 2: Determine the value of $k$ if the sum of the distances of a point from two perpendicular lines in a plane is 1, and its locus is $|x| + |y| = k$. Let the two perpendicular lines be the x-axis ($y=0$) and the y-axis ($x=0$). The distance of a point $(x_0, y_0)$ from the x-axis is $|y_0|$. The distance of a point $(x_0, y_0)$ from the y-axis is $|x_0|$. The sum of these distances is given as 1. Therefore, $|x_0| + |y_0| = 1$. Comparing this with the given locus $|x| + |y| = k$, it follows that $k=1$. Step 3: Determine the value of $m$ if $6x + 6y + m = 0$ is the acute angle bisector of lines $x + 2y + 4 = 0$ and $4x + 2y - 1 = 0$. Let the given lines be $L_1: x + 2y + 4 = 0$ and $L_2: 4x + 2y - 1 = 0$. The equations of the angle bisectors are given by: $$\frac{x + 2y + 4}{\sqrt{1^2 + 2^2}} = \pm \frac{4x + 2y - 1}{\sqrt{4^2 + 2^2}}$$ $$\frac{x + 2y + 4}{\sqrt{5}} = \pm \frac{4x + 2y - 1}{\sqrt{16 + 4}}$$ $$\frac{x + 2y + 4}{\sqrt{5}} = \pm \frac{4x + 2y - 1}{\sqrt{20}}$$ $$\frac{x + 2y + 4}{\sqrt{5}} = \pm \frac{4x + 2y - 1}{2\sqrt{5}}$$ Multiplying both sides by $2\sqrt{5}$: $$2(x + 2y + 4) = \pm (4x + 2y - 1)$$ To determine which sign corresponds to the acute angle bisector, we check the sign of $A_1A_2 + B_1B_2$. Here, $A_1=1, B_1=2$ and $A_2=4, B_2=2$. $A_1A_2 + B_1B_2 = (1)(4) + (2)(2) = 4 + 4 = 8$. Since $A_1A_2 + B_1B_2 > 0$, the negative sign in the bisector equation corresponds to the acute angle bisector. $$2(x + 2y + 4) = -(4x + 2y - 1)$$ $$2x + 4y + 8 = -4x - 2y + 1$$ $$6x + 6y + 7 = 0$$ This is the equation of the acute angle bisector. Comparing this with the given equation $6x + 6y + m = 0$, we find $m=7$. Step 4: Determine the area of the triangle formed by the lines $y^2 - 9xy + 18x^2 = 0$ and $y = 6$. The homogeneous equation $y^2 - 9xy + 18x^2 = 0$ represents two lines passing through the origin. Factoring the quadratic equation: $$y^2 - 9xy + 18x^2 = 0$$ $$(y - 3x)(y - 6x) = 0$$ This gives two lines: $y = 3x$ and $y = 6x$. The third line forming the triangle is $y = 6$. The vertices of the triangle are the intersection points of these lines: 1. Intersection of $y=3x$ and $y=6$: $6 = 3x \implies x = 2$. So, vertex $V_1 = (2, 6)$. 2. Intersection of $y=6x$ and $y=6$: $6 = 6x \implies x = 1$. So, vertex $V_2 = (1, 6)$. 3. Intersection of $y=3x$ and $y=6x$: $3x = 6x \implies 3x = 0 \implies x = 0$. Since $y=3x$, $y=0$. So, vertex $V_3 = (0, 0)$. The base of the triangle can be taken along the line $y=6$, connecting $V_1(2,6)$ and $V_2(1,6)$. The length of the base $b = |2 - 1| = 1$. The height of the triangle is the perpendicular distance from the vertex $V_3(0,0)$ to the line $y=6$. The height $h = |6 - 0| = 6$. The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. $$\text{Area} = \frac{1}{2} \times 1 \times 6 = 3$$