Foot of perpendicular, Image of a point and Reflexive properties Questions in English

(C) Let the foot of the perpendicular from the origin $(0,0)$ to the variable line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(x_1, y_1)$.
Since $P(x_1, y_1)$ lies on the line $\frac{x}{a} + \frac{y}{b} = 1$,we have $\frac{x_1}{a} + \frac{y_1}{b} = 1$.
The slope of the line $\frac{x}{a} + \frac{y}{b} = 1$ is $m_1 = -\frac{b}{a}$.
The slope of the line segment $OP$ is $m_2 = \frac{y_1}{x_1}$.
Since $OP$ is perpendicular to the line,$m_1 \times m_2 = -1$,so $(-\frac{b}{a}) \times (\frac{y_1}{x_1}) = -1$,which implies $\frac{y_1}{x_1} = \frac{a}{b}$,or $\frac{x_1}{a} = \frac{y_1}{b} = k$.
Substituting this into the line equation: $k + k = 1 \Rightarrow k = \frac{1}{2}$.
Thus,$x_1 = \frac{a}{2}$ and $y_1 = \frac{b}{2}$,which means $a = 2x_1$ and $b = 2y_1$.
Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$,we substitute $a$ and $b$: $\frac{1}{(2x_1)^2} + \frac{1}{(2y_1)^2} = \frac{1}{4}$.
$\frac{1}{4x_1^2} + \frac{1}{4y_1^2} = \frac{1}{4}$ $\Rightarrow \frac{1}{x_1^2} + \frac{1}{y_1^2} = 1$ $\Rightarrow x_1^2 + y_1^2 = x_1^2 y_1^2$.
Alternatively,using the perpendicular distance formula $d = \frac{|c|}{\sqrt{A^2 + B^2}}$ from $(0,0)$ to $\frac{x}{a} + \frac{y}{b} - 1 = 0$:
$d = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \sqrt{x_1^2 + y_1^2}$.
Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$,we have $\sqrt{x_1^2 + y_1^2} = \sqrt{4} = 2$.
Thus,$x_1^2 + y_1^2 = 4$,which is a circle with radius $2$.

(D) Mid-point of $P(2, 3)$ and $Q(4, 5)$ is $(\frac{2+4}{2}, \frac{3+5}{2}) = (3, 4)$.
Slope of $PQ = \frac{5-3}{4-2} = \frac{2}{2} = 1$.
Since the line $L$ is the perpendicular bisector of $PQ$,the slope of line $L$ is $m = -1$.
The equation of line $L$ passing through $(3, 4)$ with slope $-1$ is $y - 4 = -1(x - 3) \Rightarrow x + y - 7 = 0$.
Let the image of point $R(0, 0)$ be $S(x_1, y_1)$.
The mid-point of $RS$ is $(\frac{x_1}{2}, \frac{y_1}{2})$,which lies on line $L$,so $\frac{x_1}{2} + \frac{y_1}{2} - 7 = 0 \Rightarrow x_1 + y_1 = 14$.
The slope of $RS$ is $\frac{y_1}{x_1}$. Since $RS \perp L$,the slope of $RS$ must be $1$ (negative reciprocal of $-1$).
Thus,$\frac{y_1}{x_1} = 1 \Rightarrow x_1 = y_1$.
Substituting $x_1 = y_1$ into $x_1 + y_1 = 14$,we get $2x_1 = 14 \Rightarrow x_1 = 7$ and $y_1 = 7$.
Therefore,the image of $R$ is $(7, 7)$.

(C) Let $P(0, 1)$ be the given point and $L: 2x - y = 0$ be the line.
The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ in the line $ax + by + c = 0$ is given by $\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
Substituting the values: $\frac{x_1 - 0}{2} = \frac{y_1 - 1}{-1} = -2 \frac{2(0) - 1(1)}{2^2 + (-1)^2} = -2 \frac{-1}{5} = \frac{2}{5}$.
Thus,$x_1 = 2 \times \frac{2}{5} = \frac{4}{5}$ and $y_1 = 1 - \frac{2}{5} = \frac{3}{5}$.
So,Statement $1$ is true.
By definition,the image of a point in a line is the point such that the line is the perpendicular bisector of the segment joining the point and its image. This implies the points lie on opposite sides and are equidistant from the line. Thus,Statement $2$ is true and is the correct explanation for Statement $1$.

(B) Given the curve $x^{2}+2xy-3y^{2}=0$.
Differentiating with respect to $x$: $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$.
At point $(2,2)$: $2(2) + 2(2) + 2(2)\frac{dy}{dx} - 6(2)\frac{dy}{dx} = 0$.
$4 + 4 + 4\frac{dy}{dx} - 12\frac{dy}{dx} = 0 \implies 8 - 8\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = 1$.
The slope of the tangent at $(2,2)$ is $m_{T} = 1$.
The slope of the normal at $(2,2)$ is $m_{N} = -\frac{1}{m_{T}} = -1$.
The equation of the normal at $(2,2)$ is $(y - 2) = -1(x - 2) \implies y - 2 = -x + 2 \implies x + y - 4 = 0$.
The perpendicular distance from the origin $(0,0)$ to the line $x + y - 4 = 0$ is given by $d = \frac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}$.
$d = \frac{|0 + 0 - 4|}{\sqrt{1^{2} + 1^{2}}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

(A) Using the section formula,the coordinates of the point $P$ that divides the line segment joining $A(1,0)$ and $B(2,3)$ in the ratio $1:n$ are:
$P = \left(\frac{n(1)+1(2)}{1+n}, \frac{n(0)+1(3)}{1+n}\right) = \left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$
The slope of the line segment $AB$ is $m = \frac{3-0}{2-1} = 3$.
The slope of the line perpendicular to $AB$ is $m' = -\frac{1}{m} = -\frac{1}{3}$.
The equation of the line passing through $P$ with slope $m'$ is:
$y - \frac{3}{n+1} = -\frac{1}{3} \left(x - \frac{n+2}{n+1}\right)$
Multiplying by $3(n+1)$:
$3(n+1)y - 9 = -(x(n+1) - (n+2))$
$3(n+1)y - 9 = -(n+1)x + n + 2$
$(n+1)x + 3(n+1)y = n + 11$

(A) Let the origin be $O(0, 0)$ and the given point be $P(-2, 9)$.
The slope of the line segment $OP$ is $m_{1} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}$.
Since the line is perpendicular to $OP$,its slope $m_{2}$ is given by $m_{2} = -\frac{1}{m_{1}} = -\frac{1}{(-9/2)} = \frac{2}{9}$.
The equation of the line passing through $P(-2, 9)$ with slope $m_{2} = \frac{2}{9}$ is given by the point-slope form:
$y - y_{1} = m(x - x_{1})$
$y - 9 = \frac{2}{9}(x - (-2))$
$9(y - 9) = 2(x + 2)$
$9y - 81 = 2x + 4$
$2x - 9y + 85 = 0$.

(A) The right bisector of a line segment passes through the midpoint of the segment and is perpendicular to it.
The endpoints of the line segment are $A(3, 4)$ and $B(-1, 2)$.
The midpoint of $AB$ is $\left(\frac{3-1}{2}, \frac{4+2}{2}\right) = (1, 3)$.
The slope of $AB$ is $m_{AB} = \frac{2-4}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$.
The slope of the perpendicular bisector is $m = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2$.
The equation of the line passing through $(1, 3)$ with slope $-2$ is given by $(y - 3) = -2(x - 1)$.
$y - 3 = -2x + 2$
$2x + y = 5$.

(A) Let $(a, b)$ be the coordinates of the foot of the perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.
The slope of the line joining $(-1, 3)$ and $(a, b)$ is $m_1 = \frac{b - 3}{a + 1}$.
The slope of the given line $3x - 4y - 16 = 0$ is $m_2 = \frac{3}{4}$.
Since the two lines are perpendicular,$m_1 \times m_2 = -1$.
$\therefore \left(\frac{b - 3}{a + 1}\right) \times \left(\frac{3}{4}\right) = -1$
$\Rightarrow 3b - 9 = -4a - 4$
$\Rightarrow 4a + 3b = 5$ ..... $(1)$
Since the point $(a, b)$ lies on the line $3x - 4y - 16 = 0$,we have:
$3a - 4b = 16$ ..... $(2)$
Solving equations $(1)$ and $(2)$:
Multiply $(1)$ by $4$ and $(2)$ by $3$:
$16a + 12b = 20$
$9a - 12b = 48$
Adding these,$25a = 68 \Rightarrow a = \frac{68}{25}$.
Substituting $a$ in $(1)$: $4(\frac{68}{25}) + 3b = 5$ $\Rightarrow \frac{272}{25} + 3b = 5$ $\Rightarrow 3b = 5 - \frac{272}{25} = \frac{125 - 272}{25} = -\frac{147}{25}$ $\Rightarrow b = -\frac{49}{25}$.
Thus,the coordinates of the foot of the perpendicular are $\left(\frac{68}{25}, -\frac{49}{25}\right)$.

(A) The given equation of the line is $y = mx + c$.
The perpendicular from the origin $(0, 0)$ meets the line at the point $P(-1, 2)$.
The slope of the line segment joining $(0, 0)$ and $(-1, 2)$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$.
Since the line $y = mx + c$ is perpendicular to this segment,the product of their slopes must be $-1$.
Therefore,$m \times (-2) = -1$,which gives $m = \frac{1}{2}$.
Since the point $(-1, 2)$ lies on the line $y = mx + c$,it must satisfy the equation:
$2 = m(-1) + c$
Substituting $m = \frac{1}{2}$:
$2 = \frac{1}{2}(-1) + c$
$2 = -\frac{1}{2} + c$
$c = 2 + \frac{1}{2} = \frac{5}{2}$.
Thus,the values are $m = \frac{1}{2}$ and $c = \frac{5}{2}$.

(A) Let $Q(h, k)$ be the image of the point $P(1, 2)$ in the line $x - 3y + 4 = 0$ ... $(1)$.
The line $(1)$ is the perpendicular bisector of the line segment $PQ$.
The slope of the line $x - 3y + 4 = 0$ is $m_1 = \frac{1}{3}$.
The slope of the line $PQ$ is $m_2 = \frac{k - 2}{h - 1}$.
Since $PQ$ is perpendicular to the line $(1)$,$m_1 \times m_2 = -1$,so $\frac{1}{3} \times \frac{k - 2}{h - 1} = -1$,which gives $k - 2 = -3(h - 1)$,or $3h + k = 5$ ... $(2)$.
The midpoint of $PQ$ is $M = \left(\frac{h + 1}{2}, \frac{k + 2}{2}\right)$.
Since $M$ lies on the line $(1)$,$\frac{h + 1}{2} - 3\left(\frac{k + 2}{2}\right) + 4 = 0$,which simplifies to $h + 1 - 3k - 6 + 8 = 0$,or $h - 3k = -3$ ... $(3)$.
Solving equations $(2)$ and $(3)$:
Multiply $(2)$ by $3$: $9h + 3k = 15$.
Adding to $(3)$: $(9h + 3k) + (h - 3k) = 15 - 3$,so $10h = 12$,$h = \frac{6}{5}$.
Substituting $h$ into $(3)$: $\frac{6}{5} - 3k = -3$,so $3k = \frac{6}{5} + 3 = \frac{21}{5}$,$k = \frac{7}{5}$.
Thus,the image is $\left(\frac{6}{5}, \frac{7}{5}\right)$.

(A) The equation of the given line is $x+3y=7$ ... $(1)$.
Let point $B(a,b)$ be the image of point $A(3,8)$.
Accordingly,line $(1)$ is the perpendicular bisector of $AB$.
The slope of $AB = \frac{b-8}{a-3}$,while the slope of line $(1) = -\frac{1}{3}$.
Since line $(1)$ is perpendicular to $AB$,we have $\left(\frac{b-8}{a-3}\right) \times \left(-\frac{1}{3}\right) = -1$.
This simplifies to $\frac{b-8}{3a-9} = 1$,which gives $b-8 = 3a-9$,or $3a-b = 1$ ... $(2)$.
The mid-point of $AB$ is $\left(\frac{a+3}{2}, \frac{b+8}{2}\right)$.
Since this mid-point lies on line $(1)$,we have $\left(\frac{a+3}{2}\right) + 3\left(\frac{b+8}{2}\right) = 7$.
Multiplying by $2$,we get $a+3 + 3b+24 = 14$,which simplifies to $a+3b = -13$ ... $(3)$.
Solving equations $(2)$ and $(3)$ simultaneously: From $(2)$,$b = 3a-1$. Substituting into $(3)$,$a + 3(3a-1) = -13$,so $a + 9a - 3 = -13$,$10a = -10$,$a = -1$.
Then $b = 3(-1)-1 = -4$.
Thus,the image of the point is $(-1,-4)$.

(A) The equations of the given lines are:
$2x - 3y + 4 = 0$ $(1)$
$3x + 4y - 5 = 0$ $(2)$
$6x - 7y + 8 = 0$ $(3)$
The person is standing at the junction of the paths represented by lines $(1)$ and $(2)$.
Solving equations $(1)$ and $(2)$ by multiplying $(1)$ by $4$ and $(2)$ by $3$:
$8x - 12y + 16 = 0$
$9x + 12y - 15 = 0$
Adding these,we get $17x + 1 = 0$,so $x = -\frac{1}{17}$.
Substituting $x$ into $(1)$: $2(-\frac{1}{17}) - 3y + 4 = 0 \implies -\frac{2}{17} + 4 = 3y \implies \frac{66}{17} = 3y \implies y = \frac{22}{17}$.
Thus,the person is at point $(-\frac{1}{17}, \frac{22}{17})$.
To reach path $(3)$ in the least time,the person must walk along the perpendicular line to $(3)$ from the point $(-\frac{1}{17}, \frac{22}{17})$.
The slope of line $(3)$ is $m = \frac{6}{7}$.
The slope of the perpendicular line is $m' = -\frac{1}{m} = -\frac{7}{6}$.
The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ is:
$(y - \frac{22}{17}) = -\frac{7}{6}(x + \frac{1}{17})$
$6(17y - 22) = -7(17x + 1)$
$102y - 132 = -119x - 7$
$119x + 102y = 125$.

(D) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.
The slope of the line segment $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
The slope of the perpendicular bisector is $m = -\frac{1}{m_{PQ}} = k-1$.
The equation of the perpendicular bisector passing through $M$ with slope $m$ is $y - \frac{7}{2} = (k-1)\left(x - \frac{k+1}{2}\right)$.
The $y$-intercept is the value of $y$ when $x=0$. Given the $y$-intercept is $-4$,we substitute $x=0$ and $y=-4$ into the equation:
$-4 - \frac{7}{2} = (k-1)\left(0 - \frac{k+1}{2}\right)$
$-\frac{15}{2} = (k-1)\left(-\frac{k+1}{2}\right)$
$\frac{15}{2} = \frac{(k-1)(k+1)}{2}$
$15 = k^2 - 1$
$k^2 = 16$
$k = \pm 4$.
Since $-4$ is an option,the correct value is $-4$.

(C) The equation of the line $L$ with $x$-intercept $3$ and $y$-intercept $1$ is given by $\frac{x}{3} + \frac{y}{1} = 1$,which simplifies to $x + 3y - 3 = 0$.
To find the image $(x', y')$ of the point $(x_0, y_0) = (-1, -4)$ in the line $ax + by + c = 0$,we use the formula $\frac{x' - x_0}{a} = \frac{y' - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
Substituting the values: $\frac{x' - (-1)}{1} = \frac{y' - (-4)}{3} = -2 \frac{1(-1) + 3(-4) - 3}{1^2 + 3^2}$.
$\frac{x' + 1}{1} = \frac{y' + 4}{3} = -2 \frac{-1 - 12 - 3}{1 + 9} = -2 \frac{-16}{10} = \frac{32}{10} = \frac{16}{5}$.
For $x'$: $x' + 1 = \frac{16}{5} \Rightarrow x' = \frac{16}{5} - 1 = \frac{11}{5}$.
For $y'$: $\frac{y' + 4}{3} = \frac{16}{5}$ $\Rightarrow y' + 4 = \frac{48}{5}$ $\Rightarrow y' = \frac{48}{5} - 4 = \frac{28}{5}$.
Thus,the image is $\left(\frac{11}{5}, \frac{28}{5}\right)$.

(D) Let the image of the point $P(3,5)$ in the line $x-y+1=0$ be $P'(x,y)$.
Using the formula for the image of a point $(x_1, y_1)$ in the line $ax+by+c=0$:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2 \left( \frac{ax_1+by_1+c}{a^2+b^2} \right)$
Substituting the values $x_1=3, y_1=5, a=1, b=-1, c=1$:
$\frac{x-3}{1} = \frac{y-5}{-1} = -2 \left( \frac{3-5+1}{1^2+(-1)^2} \right)$
$\frac{x-3}{1} = \frac{y-5}{-1} = -2 \left( \frac{-1}{2} \right) = 1$
So,$x-3=1 \implies x=4$ and $y-5=-1 \implies y=4$.
The image point is $(4,4)$.
Now,check which option is satisfied by the point $(4,4)$:
For option $D$: $(4-2)^2 + (4-4)^2 = 2^2 + 0^2 = 4$.
Thus,the point $(4,4)$ lies on the circle $(x-2)^2 + (y-4)^2 = 4$.

(B) Given $A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$.
The image of $A$ in the $y$-axis is $B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$.
The image of $B$ in the $x$-axis is $C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$.
The area of $\triangle ACD$ with vertices $A\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$,$C\left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$,and $D(3 \cos \theta, a \sin \theta)$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} \left| x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C) \right|$
Substituting the coordinates:
$\text{Area} = \frac{1}{2} \left| \frac{3}{\sqrt{a}}(-\sqrt{a} - a \sin \theta) - \frac{3}{\sqrt{a}}(a \sin \theta - \sqrt{a}) + 3 \cos \theta(\sqrt{a} - (-\sqrt{a})) \right|$
$\text{Area} = \frac{1}{2} \left| -3 - 3\sqrt{a} \sin \theta - 3\sqrt{a} \sin \theta + 3 + 6\sqrt{a} \cos \theta \right|$
$\text{Area} = \frac{1}{2} \left| 6\sqrt{a} \cos \theta - 6\sqrt{a} \sin \theta \right| = 3\sqrt{a} |\cos \theta - \sin \theta|$
Since $D$ is in the fourth quadrant,$\theta \in (\frac{3\pi}{2}, 2\pi)$,so $\cos \theta > 0$ and $\sin \theta < 0$. Thus,$(\cos \theta - \sin \theta) > 0$.
The maximum value of $(\cos \theta - \sin \theta)$ is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$.
Therefore,$\text{Max Area} = 3\sqrt{a} \cdot \sqrt{2} = 12$.
$\sqrt{2a} = 4 \implies 2a = 16 \implies a = 8$.

(A) Let $P'(2, -3)$ be the reflection of $P(2, 3)$ across the $x$-axis $(y=0)$.
Since the ray reflects at $A$ on the $x$-axis,$P', A,$ and $Q$ are collinear.
The equation of line $P'Q$ passing through $(2, -3)$ and $(5, 4)$ is:
$y - (-3) = \frac{4 - (-3)}{5 - 2}(x - 2)$
$y + 3 = \frac{7}{3}(x - 2) \implies 3y + 9 = 7x - 14 \implies 7x - 3y = 23$.
At point $A$,$y = 0$,so $7x = 23 \implies x = \frac{23}{7}$. Thus,$A = (\frac{23}{7}, 0)$.
$R$ divides $AQ$ in ratio $2:1$. $A = (\frac{23}{7}, 0)$ and $Q = (5, 4)$.
$R = (\frac{2(5) + 1(23/7)}{2+1}, \frac{2(4) + 1(0)}{2+1}) = (\frac{10 + 23/7}{3}, \frac{8}{3}) = (\frac{93/7}{3}, \frac{8}{3}) = (\frac{31}{7}, \frac{8}{3})$.
The bisector of $\angle PAQ$ is the line perpendicular to the $x$-axis passing through $A$ (since the angle of incidence equals the angle of reflection). This is the line $x = \frac{23}{7}$.
The foot of the perpendicular $M$ from $R(\frac{31}{7}, \frac{8}{3})$ to the line $x = \frac{23}{7}$ is $M(\frac{23}{7}, \frac{8}{3})$.
Thus,$\alpha = \frac{23}{7}$ and $\beta = \frac{8}{3}$.
$7\alpha + 3\beta = 7(\frac{23}{7}) + 3(\frac{8}{3}) = 23 + 8 = 31$.

(C) $\triangle ABC$ is a right-angled triangle with $\angle ABC = 90^{\circ}$.
The circumcenter of $\triangle ABC$ is the midpoint of the hypotenuse $AC$,denoted as $M$.
By the properties of reflection,$AB$ is the perpendicular bisector of $PP_1$,and $BC$ is the perpendicular bisector of $PP_2$.
Since $AB \perp BC$ at $B$,the point $B$ is equidistant from $P, P_1,$ and $P_2$ because $BP = BP_1$ and $BP = BP_2$.
Thus,$B$ is the circumcenter of $\triangle P_1PP_2$.
In $\triangle ABC$,$M$ is the midpoint of $AC$,so $BM = AM = MC = \frac{AC}{2}$ (property of the median to the hypotenuse in a right triangle).
Therefore,the distance between the circumcenters $B$ and $M$ is $BM = \frac{AC}{2}$.

(C) Let $\angle PAB = \theta$ and $\angle PAC = \phi$.
Since $Q$ is the reflection of $P$ in $AB$,we have $AQ = AP$ and $\angle QAB = \angle PAB = \theta$.
Since $R$ is the reflection of $P$ in $AC$,we have $AR = AP$ and $\angle RAC = \angle PAC = \phi$.
Given that $Q, A, R$ are collinear,the angle $\angle QAR = 180^{\circ}$.
From the figure,$\angle QAR = \angle QAB + \angle BAC + \angle RAC = \theta + (\theta + \phi) + \phi = 2(\theta + \phi) = 180^{\circ}$.
Thus,$\theta + \phi = 90^{\circ}$.
Since $\angle BAC = \theta + \phi$,we have $\angle BAC = 90^{\circ}$.

(B) The incident ray passes through the origin $(0, 0)$ with slope $m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Its equation is $y = \frac{1}{\sqrt{3}}x$.
The intersection point $P$ of this ray with the line $x + y = 1$ is found by substituting $y = \frac{x}{\sqrt{3}}$ into $x + y = 1$:
$x + \frac{x}{\sqrt{3}} = 1$ $\Rightarrow x(1 + \frac{1}{\sqrt{3}}) = 1$ $\Rightarrow x = \frac{\sqrt{3}}{\sqrt{3} + 1}$.
Thus,$P = \left(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1}\right)$.
The line $x + y = 1$ has slope $m_1 = -1$,so its angle is $135^{\circ}$.
The incident ray makes $30^{\circ}$ with the $x$-axis. The angle of incidence with the normal is $135^{\circ} - 30^{\circ} = 105^{\circ}$ (relative to the line). The reflected ray makes an angle of $135^{\circ} + (135^{\circ} - 30^{\circ}) = 240^{\circ}$ or $60^{\circ}$ with the positive $x$-axis.
The slope of the reflected ray is $\tan 60^{\circ} = \sqrt{3}$.
The equation of the reflected ray passing through $P$ is $y - \frac{1}{\sqrt{3} + 1} = \sqrt{3}(x - \frac{\sqrt{3}}{\sqrt{3} + 1})$.
Setting $y = 0$ to find the $x$-intercept $Q$:
$-\frac{1}{\sqrt{3} + 1} = \sqrt{3}x - \frac{3}{\sqrt{3} + 1}$ $\Rightarrow \sqrt{3}x = \frac{3 - 1}{\sqrt{3} + 1} = \frac{2}{\sqrt{3} + 1}$.
$x = \frac{2}{\sqrt{3}(\sqrt{3} + 1)} = \frac{2}{3 + \sqrt{3}}$.

(C) To find the reflection of point $B(7,2)$ about the line $2x+y-6=0$,let the reflected point be $B'(x', y')$.
Using the reflection formula $\frac{x'-7}{2} = \frac{y'-2}{1} = -2 \left( \frac{2(7)+1(2)-6}{2^2+1^2} \right)$,
$\frac{x'-7}{2} = \frac{y'-2}{1} = -2 \left( \frac{14+2-6}{5} \right) = -2 \left( \frac{10}{5} \right) = -4$.
Thus,$x'-7 = -8 \implies x' = -1$ and $y'-2 = -4 \implies y' = -2$.
So,$B' = (-1, -2)$.
The incident ray passes through $A(3, 10)$ and $B'(-1, -2)$.
The slope of the incident ray is $m = \frac{10 - (-2)}{3 - (-1)} = \frac{12}{4} = 3$.
The equation of the incident ray is $y - 10 = 3(x - 3) \implies y - 10 = 3x - 9 \implies 3x - y + 1 = 0$.
Comparing this with $ax + by + 1 = 0$,we get $a = 3$ and $b = -1$.
Therefore,$a^2 + b^2 + 3ab = (3)^2 + (-1)^2 + 3(3)(-1) = 9 + 1 - 9 = 1$.

(D) Let $G$ be the centroid of the triangle formed by vertices $(1,3), (3,1)$ and $(2,4)$.
$G = \left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right) = \left(2, \frac{8}{3}\right)$.
Let $(\alpha, \beta)$ be the image of $G$ with respect to the line $x+2y-2=0$.
The formula for the image $(x', y')$ of a point $(x_0, y_0)$ in the line $ax+by+c=0$ is $\frac{x'-x_0}{a} = \frac{y'-y_0}{b} = -2\frac{ax_0+by_0+c}{a^2+b^2}$.
Substituting the values: $\frac{\alpha-2}{1} = \frac{\beta-8/3}{2} = -2\frac{2+2(8/3)-2}{1^2+2^2}$.
$\frac{\alpha-2}{1} = \frac{\beta-8/3}{2} = -2\frac{16/3}{5} = -\frac{32}{15}$.
Thus,$\alpha = 2 - \frac{32}{15} = -\frac{2}{15}$ and $\beta = \frac{8}{3} - \frac{64}{15} = \frac{40-64}{15} = -\frac{24}{15}$.
Finally,$15(\alpha-\beta) = 15\left(-\frac{2}{15} - (-\frac{24}{15})\right) = 15\left(\frac{22}{15}\right) = 22$.

(A) First,find the vertices of the triangle by solving the equations of the lines pairwise:
$1$) $7x-6y+3=0$ and $x+2y-31=0$ intersect at $A(9, 11)$.
$2$) $7x-6y+3=0$ and $9x-2y-19=0$ intersect at $B(3, 4)$.
$3$) $x+2y-31=0$ and $9x-2y-19=0$ intersect at $C(5, 13)$.
The centroid $G$ of $\Delta ABC$ is given by $\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) = \left(\frac{17}{3}, \frac{28}{3}\right)$.
Let $(h, k)$ be the image of $G\left(\frac{17}{3}, \frac{28}{3}\right)$ in the line $3x+6y-53=0$. The formula for the image $(h, k)$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$.
Here,$a=3, b=6, c=-53, x_1=\frac{17}{3}, y_1=\frac{28}{3}$.
$\frac{h-17/3}{3} = \frac{k-28/3}{6} = -2\frac{3(17/3)+6(28/3)-53}{3^2+6^2} = -2\frac{17+56-53}{9+36} = -2\frac{20}{45} = -2\frac{4}{9} = -\frac{8}{9}$.
$h - \frac{17}{3} = 3 \times (-\frac{8}{9}) = -\frac{8}{3} \implies h = \frac{17-8}{3} = 3$.
$k - \frac{28}{3} = 6 \times (-\frac{8}{9}) = -\frac{16}{3} \implies k = \frac{28-16}{3} = 4$.
Thus,$h^2+k^2+hk = 3^2+4^2+(3)(4) = 9+16+12 = 37$.

(A) The given equation of the family of lines is $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$.
Rearranging the terms,we get $(x+2y-5) + \lambda(3x+7y-17) = 0$.
This family of lines passes through the intersection of $x+2y-5=0$ and $3x+7y-17=0$.
Solving these equations: $x = 5-2y$,substituting into the second: $3(5-2y)+7y-17=0 \implies 15-6y+7y-17=0 \implies y=2$.
Thus,$x = 5-2(2) = 1$. The fixed point is $P(1, 2)$.
The line $L$ that is farthest from the origin $(0, 0)$ is the line perpendicular to the vector $\vec{OP}$,where $O$ is the origin.
The slope of $OP$ is $m_{OP} = \frac{2-0}{1-0} = 2$.
The slope of line $L$ is $m_L = -\frac{1}{m_{OP}} = -\frac{1}{2}$.
The equation of line $L$ passing through $P(1, 2)$ is $y-2 = -\frac{1}{2}(x-1) \implies 2y-4 = -x+1 \implies x+2y-5=0$.
We need to find the distance $d$ of line $L$ from the point $Q(3, 6)$.
$d = \frac{|1(3) + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3+12-5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$.
Therefore,$d^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$.

(D) The polar coordinates of $P$ are given as $\left(2, \frac{\pi}{6}\right)$.
If $Q$ is the reflection (image) of $P$ about the $X$-axis,the radial distance $r$ remains the same,and the angle $\theta$ changes to $-\theta$.
Therefore,the coordinates of $Q$ are $\left(2, -\frac{\pi}{6}\right)$.
To express the angle in the standard range $[0, 2\pi)$,we add $2\pi$ to the angle:
$-\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$.
Thus,the polar coordinates of $Q$ are $\left(2, \frac{11\pi}{6}\right)$.

(D) The slope of line segment $AB$ is $m_{AB} = \frac{-5-3}{6-(-2)} = \frac{-8}{8} = -1$.
Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ must satisfy $m \times m_{AB} = -1$,so $m = 1$.
The midpoint of $AB$ is $M = \left( \frac{-2+6}{2}, \frac{3-5}{2} \right) = (2, -1)$.
The equation of the line with slope $m=1$ passing through $(2, -1)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - (-1) = 1(x - 2)$,which simplifies to $y + 1 = x - 2$,or $x - y = 3$.

(C) Let the origin be $O(0, 0)$ and the foot of the perpendicular be $N(3, -4)$.
The slope of the line segment $ON$ is $m_{ON} = \frac{-4 - 0}{3 - 0} = -\frac{4}{3}$.
Since the line $L$ is perpendicular to $ON$,the slope of line $L$ $(m_L)$ is given by $m_L \times m_{ON} = -1$.
$m_L \times (-\frac{4}{3}) = -1 \Rightarrow m_L = \frac{3}{4}$.
The line $L$ passes through the point $N(3, -4)$ and has a slope $m_L = \frac{3}{4}$.
Using the point-slope form,the equation of the line $L$ is $y - y_1 = m(x - x_1)$.
$y - (-4) = \frac{3}{4}(x - 3)$
$4(y + 4) = 3(x - 3)$
$4y + 16 = 3x - 9$
$3x - 4y - 25 = 0$.

(D) Let the foot of the perpendicular be $(h, k)$.
Since $(h, k)$ lies on the line $3x - y - 1 = 0$,we have $3h - k - 1 = 0 \implies k = 3h - 1$ ... $(i)$.
The slope of the given line $3x - y - 1 = 0$ is $m_1 = 3$.
The slope of the line segment joining $(-2, 3)$ and $(h, k)$ is $m_2 = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$,so $3 \times \frac{k - 3}{h + 2} = -1$.
$3(k - 3) = -(h + 2) \implies 3k - 9 = -h - 2 \implies h + 3k = 7$ ... $(ii)$.
Substituting $k = 3h - 1$ from $(i)$ into $(ii)$:
$h + 3(3h - 1) = 7 \implies h + 9h - 3 = 7 \implies 10h = 10 \implies h = 1$.
Substituting $h = 1$ into $(i)$,$k = 3(1) - 1 = 2$.
Thus,the coordinates of the foot of the perpendicular are $(1, 2)$.

(A) Let the foot of the perpendicular be $(x, y)$.
Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$
Substituting the values $(x_1, y_1) = (1, 2)$ and $x - 3y + 7 = 0$:
$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 3(2) + 7}{1^2 + (-3)^2}$
$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 6 + 7}{1 + 9} = -\frac{2}{10} = -\frac{1}{5}$
Now,solve for $x$ and $y$:
$x - 1 = -\frac{1}{5} \Rightarrow x = 1 - \frac{1}{5} = \frac{4}{5}$
$y - 2 = -3 \times \left(-\frac{1}{5}\right) = \frac{3}{5} \Rightarrow y = 2 + \frac{3}{5} = \frac{13}{5}$
Thus,the foot of the perpendicular is $\left(\frac{4}{5}, \frac{13}{5}\right)$.

(B) Given point $(x_{1}, y_{1}) = (a \cos^{3} \theta, a \sin^{3} \theta)$.
Given line: $x \sec \theta + y \operatorname{cosec} \theta = a$.
Rewriting the given line in slope-intercept form:
$y \operatorname{cosec} \theta = -x \sec \theta + a$
$y = -\frac{\sec \theta}{\operatorname{cosec} \theta} x + \frac{a}{\operatorname{cosec} \theta}$
$y = -\frac{\sin \theta}{\cos \theta} x + a \sin \theta$.
Thus,the slope $m_{1} = -\frac{\sin \theta}{\cos \theta}$.
The slope $m$ of the line perpendicular to this line is $m = -\frac{1}{m_{1}} = \frac{\cos \theta}{\sin \theta}$.
The equation of the required line is $(y - y_{1}) = m(x - x_{1})$:
$y - a \sin^{3} \theta = \frac{\cos \theta}{\sin \theta} (x - a \cos^{3} \theta)$
$y \sin \theta - a \sin^{4} \theta = x \cos \theta - a \cos^{4} \theta$
$x \cos \theta - y \sin \theta = a \cos^{4} \theta - a \sin^{4} \theta$
$x \cos \theta - y \sin \theta = a (\cos^{2} \theta - \sin^{2} \theta) (\cos^{2} \theta + \sin^{2} \theta)$
Since $\cos^{2} \theta + \sin^{2} \theta = 1$ and $\cos^{2} \theta - \sin^{2} \theta = \cos 2 \theta$,we get:
$x \cos \theta - y \sin \theta = a \cos 2 \theta$.

(C) Let the reflection of the point $P(1, 1)$ be $Q(h, k)$.
The line of reflection is $x + y = 0$.
The mid-point of $PQ$ is $\left(\frac{h+1}{2}, \frac{k+1}{2}\right)$,which must lie on the line $x + y = 0$.
$\frac{h+1}{2} + \frac{k+1}{2} = 0 \Rightarrow h + k + 2 = 0 \quad \dots(i)$
The line $PQ$ is perpendicular to the line $x + y = 0$ (which has slope $-1$).
The slope of $PQ$ is $\frac{k-1}{h-1}$.
Since $PQ$ is perpendicular to $x + y = 0$,the product of their slopes is $-1$.
$\left(\frac{k-1}{h-1}\right) \times (-1) = -1$ $\Rightarrow \frac{k-1}{h-1} = 1$ $\Rightarrow k - 1 = h - 1$ $\Rightarrow k = h$.
Substituting $k = h$ into equation $(i)$:
$h + h + 2 = 0$ $\Rightarrow 2h = -2$ $\Rightarrow h = -1$.
Since $k = h$,we have $k = -1$.
Thus,the reflection of the point $(1, 1)$ along the line $y = -x$ is $(-1, -1)$.

(C) Let the given line be $L_1: x+y=4$. The slope of $L_1$ is $m_1 = -1$.
The slope of the line $L_2$ perpendicular to $L_1$ is $m_2 = -\frac{1}{m_1} = 1$.
The equation of the line $L_2$ passing through $(2,4)$ is $y-4 = 1(x-2)$,which simplifies to $y-x=2$.
To find the foot of the perpendicular,we solve the system of equations:
$x+y=4$
$y-x=2$
Adding the two equations: $2y = 6 \Rightarrow y=3$.
Substituting $y=3$ into $x+y=4$,we get $x+3=4 \Rightarrow x=1$.
Thus,the foot of the perpendicular is $(1,3)$.

(A) We know that the foot of the perpendicular $(h, k)$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by the formula:
$\frac{h-x_1}{a} = \frac{k-y_1}{b} = \frac{-(ax_1+by_1+c)}{a^2+b^2}$
Here,the point $(x_1, y_1) = (3,4)$ and the line is $2x+y-7=0$.
Thus,$a=2, b=1, c=-7$.
Substituting these values into the formula:
$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-(2(3) + 1(4) - 7)}{2^2 + 1^2}$
$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-(6+4-7)}{4+1}$
$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-3}{5}$
Now,solving for $h$:
$h-3 = 2 \times \left(\frac{-3}{5}\right) = \frac{-6}{5}$
$h = 3 - \frac{6}{5} = \frac{15-6}{5} = \frac{9}{5}$
Solving for $k$:
$k-4 = 1 \times \left(\frac{-3}{5}\right) = \frac{-3}{5}$
$k = 4 - \frac{3}{5} = \frac{20-3}{5} = \frac{17}{5}$
Therefore,the coordinates of the foot of the perpendicular are $\left(\frac{9}{5}, \frac{17}{5}\right)$.

(A) To find the shortest path from $A(-3, 4)$ to $C(0, 1)$ touching the line $L: 2x + y - 7 = 0$,we reflect point $A$ across the line $L$ to get $A'$.
Let $A'(x', y')$ be the reflection of $A(-3, 4)$ in $2x + y - 7 = 0$.
Using the formula $\frac{x' - (-3)}{2} = \frac{y' - 4}{1} = -2 \frac{2(-3) + 4 - 7}{2^2 + 1^2} = -2 \frac{-9}{5} = \frac{18}{5}$.
$x' = -3 + 2(\frac{18}{5}) = -3 + \frac{36}{5} = \frac{21}{5}$.
$y' = 4 + 1(\frac{18}{5}) = 4 + \frac{18}{5} = \frac{38}{5}$.
So,$A' = (\frac{21}{5}, \frac{38}{5})$.
The shortest path distance is the length of the line segment $A'C$.
The point $B$ is the intersection of $A'C$ and the line $2x + y - 7 = 0$.
The line $A'C$ passes through $(\frac{21}{5}, \frac{38}{5})$ and $(0, 1)$.
Slope $m = \frac{1 - 38/5}{0 - 21/5} = \frac{-33/5}{-21/5} = \frac{33}{21} = \frac{11}{7}$.
Equation of $A'C$: $y - 1 = \frac{11}{7}(x - 0) \Rightarrow 11x - 7y + 7 = 0$.
Solving $2x + y - 7 = 0$ and $11x - 7y + 7 = 0$:
$y = 7 - 2x$ $\Rightarrow 11x - 7(7 - 2x) + 7 = 0$ $\Rightarrow 11x - 49 + 14x + 7 = 0$ $\Rightarrow 25x = 42$ $\Rightarrow x = \frac{42}{25}$.
$y = 7 - 2(\frac{42}{25}) = \frac{175 - 84}{25} = \frac{91}{25}$.
So,$B = (\frac{42}{25}, \frac{91}{25})$.
Distance $AB = \sqrt{(\frac{42}{25} + 3)^2 + (\frac{91}{25} - 4)^2} = \sqrt{(\frac{117}{25})^2 + (\frac{-9}{25})^2} = \frac{1}{25} \sqrt{13689 + 81} = \frac{\sqrt{13770}}{25} = \frac{9 \sqrt{170}}{25}$.

(C) Let the line passing through $A(3,1)$ be $y - 1 = m(x - 3)$,which simplifies to $mx - y + (1 - 3m) = 0$.
The distance $d$ of this line from the origin $O(0,0)$ is given by $d = \frac{|1 - 3m|}{\sqrt{m^2 + 1}}$.
For the distance to be maximum,the line must be perpendicular to the line segment $OA$.
The slope of $OA$ is $\frac{1-0}{3-0} = \frac{1}{3}$.
Since the line is perpendicular to $OA$,its slope $m$ must satisfy $m \times \frac{1}{3} = -1$,so $m = -3$.
Substituting $m = -3$ into the line equation: $-3x - y + (1 - 3(-3)) = 0$,which gives $3x + y = 10$.
To find the intercepts $P$ and $Q$,set $y=0$ to get $x = \frac{10}{3}$ (point $P(\frac{10}{3}, 0)$) and set $x=0$ to get $y = 10$ (point $Q(0, 10)$).
The area of $\triangle OPQ = \frac{1}{2} \times |OP| \times |OQ| = \frac{1}{2} \times \frac{10}{3} \times 10 = \frac{50}{3}$ sq. units.

(C) The centroid $G$ divides the line segment joining the orthocentre $H(5,8)$ and the circumcentre $O(h,k)$ in the ratio $2:1$.
Using the section formula,we have:
$\left(\frac{2h+5}{3}, \frac{2k+8}{3}\right) = \left(3, \frac{14}{3}\right)$
Equating the coordinates:
$\frac{2h+5}{3} = 3$ $\Rightarrow 2h+5 = 9$ $\Rightarrow h = 2$
$\frac{2k+8}{3} = \frac{14}{3}$ $\Rightarrow 2k+8 = 14$ $\Rightarrow k = 3$
So,the circumcentre $O$ is $(2,3)$.
The image of the orthocentre $H(5,8)$ with respect to the line $x-y=0$ (side $BC$) lies on the circumcircle. Let the image be $(x', y')$.
Using the formula $\frac{x'-5}{1} = \frac{y'-8}{-1} = -2 \frac{5-8}{1^2+(-1)^2} = -2 \frac{-3}{2} = 3$.
$x'-5 = 3 \Rightarrow x' = 8$
$y'-8 = -3 \Rightarrow y' = 5$
The point $(8,5)$ lies on the circumcircle. The radius $R$ is the distance between the circumcentre $(2,3)$ and $(8,5)$:
$R = \sqrt{(8-2)^2 + (5-3)^2} = \sqrt{6^2 + 2^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$.
The diameter is $2R = 2(2\sqrt{10}) = 4\sqrt{10}$.

(B) Step $1$: Reflection of point $P(4,1)$ in the line $x-y=0$ (or $y=x$). The rule for reflection in $y=x$ is $(x,y) \to (y,x)$. Thus,the new point $P'$ is $(1,4)$.
Step $2$: Translation through a distance of $2$ units along the positive $X$-axis. The rule is $(x,y) \to (x+2, y)$. Thus,$P'' = (1+2, 4) = (3,4)$.
Step $3$: Projection on the $X$-axis. The projection of a point $(x,y)$ on the $X$-axis is $(x,0)$. Thus,the final point is $(3,0)$.

(D) The reflection of a point $(x, y)$ about the $X$-axis is $(x, -y)$.
Therefore,the coordinates of $B$ are $(2, -3)$.
The reflection of a point $(x, y)$ about the line $x+y=0$ is $(-y, -x)$.
Therefore,the coordinates of $C$ are $(-(-3), -(2)) = (3, -2)$.
The reflection of a point $(x, y)$ about the line $x-y=0$ is $(y, x)$.
Therefore,the coordinates of $D$ are $(-2, 3)$.
The line $AB$ passes through $(2, 3)$ and $(2, -3)$,so its equation is $x=2$.
The line $CD$ passes through $(3, -2)$ and $(-2, 3)$. The slope $m = \frac{3 - (-2)}{-2 - 3} = \frac{5}{-5} = -1$.
The equation of $CD$ is $y - 3 = -1(x + 2)$ $\Rightarrow y - 3 = -x - 2$ $\Rightarrow x + y = 1$.
Substituting $x=2$ into $x+y=1$,we get $2+y=1 \Rightarrow y=-1$.
Thus,the point of intersection is $(2, -1)$.

(B) The given line is $x \sec \theta + y \operatorname{cosec} \theta = a$,which can be written as $\frac{x}{\cos \theta} + \frac{y}{\sin \theta} = a$.
Any line perpendicular to $Ax + By + C = 0$ is of the form $Bx - Ay + k = 0$.
Thus,the line perpendicular to the given line is $x \operatorname{cosec} \theta - y \sec \theta + k = 0$,or $\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = -k$.
Since this line passes through $(a \cos^3 \theta, a \sin^3 \theta)$,we substitute these coordinates:
$\frac{a \cos^3 \theta}{\sin \theta} - \frac{a \sin^3 \theta}{\cos \theta} = -k$
$\frac{a(\cos^4 \theta - \sin^4 \theta)}{\sin \theta \cos \theta} = -k$
Using $\cos^4 \theta - \sin^4 \theta = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) = \cos 2 \theta$,we get:
$\frac{a \cos 2 \theta}{\sin \theta \cos \theta} = -k$
Substituting $-k$ back into the equation $\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = -k$:
$\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = \frac{a \cos 2 \theta}{\sin \theta \cos \theta}$
Multiplying by $\sin \theta \cos \theta$ gives:
$x \cos \theta - y \sin \theta = a \cos 2 \theta$.

(A) The given line is $x \cos \theta - y \sin \theta = 2 \cos 2 \theta$.
Comparing this with $y = mx + c$,the slope $m_1 = \frac{\cos \theta}{\sin \theta}$.
The slope of the line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{\sin \theta}{\cos \theta}$.
The equation of the line passing through $(2 \cos^3 \theta, 2 \sin^3 \theta)$ with slope $m_2$ is:
$y - 2 \sin^3 \theta = -\frac{\sin \theta}{\cos \theta} (x - 2 \cos^3 \theta)$
$y \cos \theta - 2 \sin^3 \theta \cos \theta = -x \sin \theta + 2 \cos^3 \theta \sin \theta$
$x \sin \theta + y \cos \theta = 2 \sin \theta \cos \theta (\cos^2 \theta + \sin^2 \theta)$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$x \sin \theta + y \cos \theta = 2 \sin \theta \cos \theta$
Dividing both sides by $\sin \theta \cos \theta$:
$\frac{x \sin \theta}{\sin \theta \cos \theta} + \frac{y \cos \theta}{\sin \theta \cos \theta} = \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$x \sec \theta + y \operatorname{cosec} \theta = 2$.

(B) The person is standing at the intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$.
Solving these equations,we multiply the first by $4$ and the second by $3$:
$8x - 12y + 16 = 0$
$9x + 12y - 15 = 0$
Adding these gives $17x + 1 = 0$,so $x = -\frac{1}{17}$.
Substituting $x$ into the first equation: $2(-\frac{1}{17}) - 3y + 4 = 0$ $\Rightarrow -\frac{2}{17} + 4 = 3y$ $\Rightarrow 3y = \frac{66}{17}$ $\Rightarrow y = \frac{22}{17}$.
The junction point is $A = (-\frac{1}{17}, \frac{22}{17})$.
To reach the path $6x - 7y + 8 = 0$ in the least time,the person must walk along the perpendicular line.
The slope of the given path is $m_1 = \frac{6}{7}$.
The slope of the perpendicular line is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$.
The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ is:
$y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$
$6(17y - 22) = -7(17x + 1)$
$102y - 132 = -119x - 7$
$119x + 102y - 125 = 0$.

(A) Let $B = (x_1, y_1)$. Since $3x + 4y - 18 = 0$ is the perpendicular bisector of $AB$,the midpoint $M$ of $AB$ lies on the line.
$M = (\frac{x_1+3}{2}, \frac{y_1-4}{2})$. Substituting into the line equation: $3(\frac{x_1+3}{2}) + 4(\frac{y_1-4}{2}) - 18 = 0 \implies 3x_1 + 9 + 4y_1 - 16 - 36 = 0 \implies 3x_1 + 4y_1 = 43$.
Also,the slope of $AB$ is perpendicular to the line $3x + 4y - 18 = 0$ (slope $-3/4$). So,slope of $AB = 4/3$.
$\frac{y_1+4}{x_1-3} = \frac{4}{3} \implies 3y_1 + 12 = 4x_1 - 12 \implies 4x_1 - 3y_1 = 24$.
Solving the system: $3x_1 + 4y_1 = 43$ and $4x_1 - 3y_1 = 24$. Multiplying by $3$ and $4$: $9x_1 + 12y_1 = 129$ and $16x_1 - 12y_1 = 96$. Adding gives $25x_1 = 225 \implies x_1 = 9$. Then $4(9) - 3y_1 = 24 \implies 36 - 24 = 3y_1 \implies y_1 = 4$. So $B = (9, 4)$.
The centroid $G$ is $(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}) = (\frac{3+9+6}{3}, \frac{-4+4+3}{3}) = (6, 1)$.

(A) To find the image of the line $x+y-2=0$ in the $y$-axis,we replace $x$ with $-x$ in the equation of the line.
Substituting $x = -x$ into $x+y-2=0$,we get:
$(-x)+y-2=0$
$-x+y-2=0$
Multiplying by $-1$,we get:
$x-y+2=0$
Alternatively,the line $x+y=2$ passes through points $A(2, 0)$ and $B(0, 2)$.
The image of $A(2, 0)$ in the $y$-axis is $A'(-2, 0)$,and the image of $B(0, 2)$ is $B(0, 2)$ itself.
The equation of the line passing through $A'(-2, 0)$ and $B(0, 2)$ is:
$y-0 = \frac{2-0}{0-(-2)}(x-(-2))$
$y = \frac{2}{2}(x+2)$
$y = x+2$
$x-y+2=0$

(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha + 1}{2}, \frac{\beta - 2}{2}\right)$.
Since $M$ lies on the line $2x - 3y + 5 = 0$,we have $2\left(\frac{\alpha + 1}{2}\right) - 3\left(\frac{\beta - 2}{2}\right) + 5 = 0$,which simplifies to $2\alpha - 3\beta + 18 = 0$ $(i)$.
The slope of line $AB$ is $m_1 = \frac{\beta + 2}{\alpha - 1}$. The slope of the given line $2x - 3y + 5 = 0$ is $m_2 = \frac{2}{3}$.
Since $AB$ is perpendicular to the given line,$m_1 \times m_2 = -1$,so $\left(\frac{\beta + 2}{\alpha - 1}\right) \times \frac{2}{3} = -1$,which gives $2\beta + 4 = -3\alpha + 3$,or $3\alpha + 2\beta + 1 = 0$ $(ii)$.
Solving equations $(i)$ and $(ii)$ simultaneously: Multiply $(i)$ by $2$ and $(ii)$ by $3$: $4\alpha - 6\beta + 36 = 0$ and $9\alpha + 6\beta + 3 = 0$.
Adding these gives $13\alpha + 39 = 0$,so $\alpha = -3$. Substituting $\alpha = -3$ into $(ii)$,we get $3(-3) + 2\beta + 1 = 0$,so $2\beta = 8$,which means $\beta = 4$.
Thus,$\alpha + \beta = -3 + 4 = 1$.

(C) The midpoint $M$ of $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{1+k}{2}, \frac{7}{2}\right)$.
The slope of line $PQ$ is $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
The slope of the perpendicular bisector $m'$ is given by $m \times m' = -1$,so $m' = \frac{1}{m} = -(k-1) = 1-k$.
The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)(x - \frac{1+k}{2})$.
For the $y$-intercept,we set $x = 0$ and $y = -4$:
$-4 - \frac{7}{2} = (1-k)(0 - \frac{1+k}{2})$
$-\frac{15}{2} = (1-k)(-\frac{1+k}{2})$
$15 = (1-k)(1+k)$
$15 = 1 - k^2$
$k^2 = 1 - 15 = -14$.
Wait,re-evaluating the slope calculation: $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$. Thus $m' = k-1$.
The equation is $y - \frac{7}{2} = (k-1)(x - \frac{1+k}{2})$.
Setting $x=0, y=-4$:
$-4 - \frac{7}{2} = (k-1)(-\frac{1+k}{2})$
$-\frac{15}{2} = -\frac{k^2-1}{2}$
$15 = k^2 - 1$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.
Thus,a possible value of $k$ is $-4$.

(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which has a slope $m_2 = -2$. The point of incidence is $(0, 1)$.
Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror is equal to the angle between the reflected ray and the mirror.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{7 - (-2)}{1 + 7(-2)} \right| = \left| \frac{9}{1 - 14} \right| = \left| \frac{9}{-13} \right| = \frac{9}{13}$.
$\frac{m - 7}{1 + 7m} = \frac{9}{13}$ or $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$.
Case $1$: $13(m - 7) = 9(1 + 7m)$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. This is the slope of the reflected ray.
Case $2$: $13(m - 7) = -9(1 + 7m)$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.
The equation of the incident line passing through $(0, 1)$ with slope $m = \frac{41}{38}$ is:
$y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

(B) The equation of the incident line is $x-2y-3=0$. The slope of this line is $m_1 = \frac{1}{2}$.
Since the particle is reflected in a perpendicular direction,the slope of the reflected line $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -2$.
The point of incidence is the intersection of $x-2y-3=0$ and $3x-2y-5=0$.
Subtracting the first equation from the second: $(3x-2y-5) - (x-2y-3) = 0$ $\Rightarrow 2x-2=0$ $\Rightarrow x=1$.
Substituting $x=1$ into $x-2y-3=0$,we get $1-2y-3=0$ $\Rightarrow -2y=2$ $\Rightarrow y=-1$.
The point of intersection is $(1, -1)$.
The equation of the reflected line passing through $(1, -1)$ with slope $m_2 = -2$ is:
$y - (-1) = -2(x - 1)$
$y + 1 = -2x + 2$
$2x + y - 1 = 0$.

(A) Let $l_1 \equiv 2x + 3y = 5$.
Since the line $AB$ is perpendicular to $l_1$,the slope of $l_1$ is $m_1 = -\frac{2}{3}$.
Therefore,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.
The equation of line $AB$ passing through $A\left(1, \frac{1}{3}\right)$ with slope $\frac{3}{2}$ is:
$\left(y - \frac{1}{3}\right) = \frac{3}{2}(x - 1)$
$\Rightarrow 2y - \frac{2}{3} = 3x - 3$
$\Rightarrow 3x - 2y = 3 - \frac{2}{3} = \frac{7}{3}$
$\Rightarrow 9x - 6y = 7$ $(i)$
The equation of line $l_1$ is $2x + 3y = 5$. Multiplying by $2$,we get $4x + 6y = 10$ (ii).
Adding $(i)$ and (ii): $13x = 17 \Rightarrow x = \frac{17}{13}$.
Substituting $x$ in $2x + 3y = 5$: $2\left(\frac{17}{13}\right) + 3y = 5$ $\Rightarrow 3y = 5 - \frac{34}{13} = \frac{31}{13}$ $\Rightarrow y = \frac{31}{39}$.
The intersection point $P$ (mid-point of $AB$) is $\left(\frac{17}{13}, \frac{31}{39}\right)$.
Let $B = (x_2, y_2)$. Since $P$ is the mid-point of $AB$:
$\frac{1 + x_2}{2} = \frac{17}{13}$ $\Rightarrow 1 + x_2 = \frac{34}{13}$ $\Rightarrow x_2 = \frac{21}{13}$
$\frac{1/3 + y_2}{2} = \frac{31}{39}$ $\Rightarrow \frac{1}{3} + y_2 = \frac{62}{39}$ $\Rightarrow y_2 = \frac{62}{39} - \frac{13}{39} = \frac{49}{39}$
Thus,$B = \left(\frac{21}{13}, \frac{49}{39}\right)$.