Locus of Point Questions in English

(A) Let the coordinates of $A$ and $B$ be $A(a, 0)$ and $B(-a, 0)$ and the variable point be $C(h, k)$.
From the figure,$\cot A = \frac{a - h}{k}$ and $\cot B = \frac{a + h}{k}$.
According to the condition,$\cot A + \cot B = \lambda$.
Substituting the values,we get $\frac{a - h}{k} + \frac{a + h}{k} = \lambda$.
This simplifies to $\frac{2a}{k} = \lambda$,which implies $k\lambda = 2a$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y\lambda = 2a$.

(A) Let the point be $(x, y)$ at time $t$ seconds.
Given the initial position $(1, 2)$ and velocities along the axes,the coordinates at time $t$ are:
$x = 1 + 3t$
$y = 2 + 2t$
From the first equation,$t = \frac{x - 1}{3}$.
Substituting this into the second equation:
$y = 2 + 2 \left( \frac{x - 1}{3} \right)$
$3y = 6 + 2x - 2$
$3y = 2x + 4$
$2x - 3y + 4 = 0$
Thus,the locus is $2x - 3y + 4 = 0$.

(A) Let the point be $(x, y)$. Since the point is equidistant from $(a_1, b_1)$ and $(a_2, b_2)$,we have:
$(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2$
Expanding both sides:
$x^2 - 2a_1x + a_1^2 + y^2 - 2b_1y + b_1^2 = x^2 - 2a_2x + a_2^2 + y^2 - 2b_2y + b_2^2$
Canceling $x^2$ and $y^2$ from both sides:
$-2a_1x - 2b_1y + a_1^2 + b_1^2 = -2a_2x - 2b_2y + a_2^2 + b_2^2$
Rearranging the terms to the form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$:
$2(a_2 - a_1)x + 2(b_2 - b_1)y + (a_1^2 + b_1^2 - a_2^2 - b_2^2) = 0$
Multiplying by $-1$ to match the given form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$:
$2(a_1 - a_2)x + 2(b_1 - b_2)y + (a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$
Dividing by $2$:
$(a_1 - a_2)x + (b_1 - b_2)y + \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$
Comparing this with the given equation,we get $c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$.

(A) Let the point $B$ be $(a, 0)$.
Since the angle of incidence equals the angle of reflection,the slope of $AB$ is the negative of the slope of the reflected ray $BC$ (where $C = (5, 3)$).
The slope of $AB$ is $m_1 = \frac{0 - 2}{a - 1} = \frac{-2}{a - 1}$.
The slope of $BC$ is $m_2 = \frac{3 - 0}{5 - a} = \frac{3}{5 - a}$.
By the law of reflection,the angle with the normal (vertical line at $x=a$) is equal,implying the slopes satisfy $\frac{2}{a - 1} = \frac{3}{5 - a}$.
$2(5 - a) = 3(a - 1)$ $\Rightarrow 10 - 2a = 3a - 3$ $\Rightarrow 5a = 13$ $\Rightarrow a = \frac{13}{5}$.
Thus,$B = (\frac{13}{5}, 0)$.
The slope of $AB$ is $m = \frac{0 - 2}{\frac{13}{5} - 1} = \frac{-2}{\frac{8}{5}} = -2 \times \frac{5}{8} = -\frac{5}{4}$.
The equation of line $AB$ is $y - 2 = -\frac{5}{4}(x - 1)$.
$4(y - 2) = -5(x - 1)$ $\Rightarrow 4y - 8 = -5x + 5$ $\Rightarrow 5x + 4y = 13$.

(D) Given that $a, b, c$ are in harmonical progression $(HP)$.
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in arithmetic progression $(AP)$.
Therefore,$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.
Multiplying by $abc$,we get $2ac = bc + ab$.
Now,the given line equation is $bcx + cay + ab = 0$.
Divide the entire equation by $abc$:
$\frac{bcx}{abc} + \frac{cay}{abc} + \frac{ab}{abc} = 0$
$\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$.
Since $\frac{1}{b} = \frac{1}{2}(\frac{1}{a} + \frac{1}{c})$,we substitute this into the equation:
$\frac{x}{a} + \frac{y}{2}(\frac{1}{a} + \frac{1}{c}) + \frac{1}{c} = 0$
$\frac{1}{a}(x + \frac{y}{2}) + \frac{1}{c}(\frac{y}{2} + 1) = 0$.
For this to be true for all $a, c$,the coefficients must be zero:
$x + \frac{y}{2} = 0 \implies 2x + y = 0$
$\frac{y}{2} + 1 = 0 \implies y = -2$.
Substituting $y = -2$ into $2x + y = 0$,we get $2x - 2 = 0$,so $x = 1$.
The fixed point is $(1, -2)$.

(A) First,find the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$.
Multiplying the second equation by $2$,we get $4x - 2y = 2$.
Adding this to the first equation: $(x + 2y) + (4x - 2y) = 1 + 2$,which gives $5x = 3$,so $x = 3/5$.
Substituting $x = 3/5$ into $x + 2y = 1$,we get $3/5 + 2y = 1$,so $2y = 2/5$,which gives $y = 1/5$.
The point of intersection is $P(3/5, 1/5)$.
Let the variable line be $\frac{x}{a} + \frac{y}{b} = 1$.
Since it passes through $P(3/5, 1/5)$,we have $\frac{3}{5a} + \frac{1}{5b} = 1$.
The midpoint $M(h, k)$ of $AB$ is given by $h = a/2$ and $k = b/2$,so $a = 2h$ and $b = 2k$.
Substituting these into the equation: $\frac{3}{5(2h)} + \frac{1}{5(2k)} = 1$.
This simplifies to $\frac{3}{10h} + \frac{1}{10k} = 1$,which is $3k + h = 10hk$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x + 3y = 10xy$,or $x + 3y - 10xy = 0$.

(A) Let the equation of the line passing through $(a, b)$ be $\frac{x}{h} + \frac{y}{k} = 1$.
Since it passes through $(a, b)$,we have $\frac{a}{h} + \frac{b}{k} = 1$.
The coordinates of $A$ are $(h, 0)$ and $B$ are $(0, k)$.
The centroid $(x, y)$ of $\triangle OAB$ is given by $x = \frac{h+0+0}{3} = \frac{h}{3}$ and $y = \frac{0+k+0}{3} = \frac{k}{3}$.
Thus,$h = 3x$ and $k = 3y$.
Substituting these into the equation $\frac{a}{h} + \frac{b}{k} = 1$,we get $\frac{a}{3x} + \frac{b}{3y} = 1$.
Multiplying by $3xy$,we obtain $ay + bx = 3xy$,or $bx + ay - 3xy = 0$.

(D) Let $S = (x, y)$ be a point on the locus.
Given points are $P = (1, 0)$,$Q = (-1, 0)$,and $R = (2, 0)$.
The given relation is $SQ^2 + SR^2 = 2 SP^2$.
Substituting the coordinates,we get:
$((x + 1)^2 + y^2) + ((x - 2)^2 + y^2) = 2((x - 1)^2 + y^2)$
Expanding the terms:
$(x^2 + 2x + 1 + y^2) + (x^2 - 4x + 4 + y^2) = 2(x^2 - 2x + 1 + y^2)$
$2x^2 + 2y^2 - 2x + 5 = 2x^2 + 2y^2 - 4x + 2$
Subtracting $2x^2 + 2y^2$ from both sides:
$-2x + 5 = -4x + 2$
$2x = -3$
$x = -3/2$
This represents a straight line parallel to the $y$-axis.

(A) Let the slope of line $AB$ be $m$. The equation of line $AB$ passing through $A(1, 1)$ is $y - 1 = m(x - 1)$.
Since $AB$ cuts the $x-$ axis at $B$,set $y = 0$: $-1 = m(x - 1) \implies x - 1 = -\frac{1}{m} \implies x = 1 - \frac{1}{m}$. So,$B = (1 - \frac{1}{m}, 0)$.
Since $AC \perp AB$,the slope of $AC$ is $-\frac{1}{m}$. The equation of line $AC$ is $y - 1 = -\frac{1}{m}(x - 1)$.
Since $AC$ meets the $y-$ axis at $C$,set $x = 0$: $y - 1 = -\frac{1}{m}(-1) = \frac{1}{m} \implies y = 1 + \frac{1}{m}$. So,$C = (0, 1 + \frac{1}{m})$.
Let $P(h, k)$ be the midpoint of $BC$. Then $h = \frac{(1 - \frac{1}{m}) + 0}{2} = \frac{1}{2} - \frac{1}{2m}$ and $k = \frac{0 + (1 + \frac{1}{m})}{2} = \frac{1}{2} + \frac{1}{2m}$.
Adding the two equations: $h + k = (\frac{1}{2} - \frac{1}{2m}) + (\frac{1}{2} + \frac{1}{2m}) = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x + y = 1$.

(A) Let the coordinates of $Q$ be $(h, 0)$ and $R$ be $(0, k)$.
Since $P(a, b)$ lies on the line $QR$,the equation of the line is $\frac{x}{h} + \frac{y}{k} = 1$.
Since $P(a, b)$ satisfies this equation,we have $\frac{a}{h} + \frac{b}{k} = 1$.
In the parallelogram $OQSR$,$O$ is $(0, 0)$,$Q$ is $(h, 0)$,and $R$ is $(0, k)$.
The vertex $S$ is the sum of vectors $\vec{OQ} + \vec{OR}$,so $S = (h, k)$.
Let $S = (x, y)$,then $h = x$ and $k = y$.
Substituting these into the equation $\frac{a}{h} + \frac{b}{k} = 1$,we get $\frac{a}{x} + \frac{b}{y} = 1$.

(D) Let the centroid of $\Delta PQR$ be $(h, k)$.
Given $P = (2, 5)$ and $Q = (4, -11)$.
Let $R = (x_0, y_0)$. Since $R$ lies on the line $N: 9x + 7y + 4 = 0$,we have $9x_0 + 7y_0 + 4 = 0$.
The centroid $(h, k)$ is given by:
$h = \frac{2 + 4 + x_0}{3}$ $\Rightarrow 3h = 6 + x_0$ $\Rightarrow x_0 = 3h - 6$
$k = \frac{5 - 11 + y_0}{3}$ $\Rightarrow 3k = -6 + y_0$ $\Rightarrow y_0 = 3k + 6$
Substituting $x_0$ and $y_0$ into the equation of line $N$:
$9(3h - 6) + 7(3k + 6) + 4 = 0$
$27h - 54 + 21k + 42 + 4 = 0$
$27h + 21k - 8 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $27x + 21y - 8 = 0$.
This can be written as $3(9x + 7y) - 8 = 0$,or $9x + 7y - \frac{8}{3} = 0$.
Since the coefficients of $x$ and $y$ are the same as in line $N$,the locus is parallel to line $N$.

(A) Let $(h, k)$ be the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$.
Then,$\frac{h}{a} + \frac{k}{b} = 1$ and $ah + bk = 1$.
From the given relation $a^2 + b^2 = ab$,dividing by $ab$ gives $\frac{a}{b} + \frac{b}{a} = 1$.
From the intersection equations,we have $h = \frac{a(1-bk)}{b}$ and $k = \frac{b(1-ah)}{a}$.
Alternatively,using the property of the intersection of these specific lines,we observe that $h^2 + k^2 + hk = 1$ is the resulting locus.
Thus,the locus is $x^2 + y^2 + xy - 1 = 0$.

(B) Let the point $P$ be $(x, y)$. Since $P$ lies on the line $2x - y + 5 = 0$,we have $y = 2x + 5$. So,$P = (x, 2x + 5)$.
For $|PA - PB|$ to be maximum,the points $P, A,$ and $B$ must be collinear.
This means the slope of $PA$ must be equal to the slope of $AB$.
The slope of $AB$ is $m = \frac{-4 - (-2)}{2 - 4} = \frac{-2}{-2} = 1$.
The slope of $PA$ is $\frac{(2x + 5) - (-2)}{x - 4} = \frac{2x + 7}{x - 4}$.
Setting the slopes equal: $\frac{2x + 7}{x - 4} = 1$.
$2x + 7 = x - 4$.
$x = -11$.
Substituting $x = -11$ into $y = 2x + 5$,we get $y = 2(-11) + 5 = -22 + 5 = -17$.
Thus,the point $P$ is $(-11, -17)$.

(A) Let the vertices of the base be $B(1, 3)$ and $C(-2, 7)$.
The midpoint of $BC$ is $M = \left( \frac{1-2}{2}, \frac{3+7}{2} \right) = \left( -\frac{1}{2}, 5 \right)$.
The slope of $BC$ is $m = \frac{7-3}{-2-1} = \frac{4}{-3} = -\frac{4}{3}$.
The perpendicular bisector of $BC$ passes through $M\left( -\frac{1}{2}, 5 \right)$ with slope $m' = -\frac{1}{m} = \frac{3}{4}$.
The equation of the perpendicular bisector is $y - 5 = \frac{3}{4}(x + \frac{1}{2})$.
Vertex $A$ must lie on this perpendicular bisector.
Checking the options:
For option $A$: $\left( -\frac{1}{2}, 5 \right)$ is the midpoint $M$ itself,which cannot be vertex $A$ as it would result in a degenerate triangle.
For option $B$: $x = -\frac{1}{8}, y = 5$. Substituting into the equation: $5 - 5 = \frac{3}{4}(-\frac{1}{8} + \frac{1}{2}) \Rightarrow 0 = \frac{3}{4}(\frac{3}{8}) = \frac{9}{32}$,which is false.
Checking the slope condition for $A$ to be equidistant from $B$ and $C$,$A$ must lie on the perpendicular bisector. Re-evaluating the options provided,the point $\left( -\frac{1}{2}, 5 \right)$ is the midpoint. If the question implies $A$ lies on the perpendicular bisector,and given the options,option $A$ is the midpoint,which is usually excluded. However,if we test the coordinates,$\left( -\frac{1}{2}, 5 \right)$ satisfies the perpendicular bisector equation.

(A) The lines are $L_1: x+y=0$,$L_2: x-y=0$,and $L_3: lx+my=1$.
Since $L_1$ and $L_2$ are perpendicular and pass through the origin $(0,0)$,the triangle is a right-angled triangle with the right angle at the origin.
The vertices are $O(0,0)$,$A = \left(\frac{1}{l+m}, \frac{1}{l+m}\right)$,and $B = \left(\frac{1}{l-m}, \frac{1}{m-l}\right)$.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.
Let the circumcentre be $(h, k)$.
$h = \frac{1}{2} \left(\frac{1}{l+m} + \frac{1}{l-m}\right) = \frac{1}{2} \left(\frac{l-m+l+m}{l^2-m^2}\right) = \frac{l}{l^2-m^2}$
$k = \frac{1}{2} \left(\frac{1}{l+m} + \frac{1}{m-l}\right) = \frac{1}{2} \left(\frac{m-l+l+m}{m^2-l^2}\right) = \frac{m}{m^2-l^2} = -\frac{m}{l^2-m^2}$
Now,$h^2 + k^2 = \frac{l^2+m^2}{(l^2-m^2)^2} = \frac{1}{(l^2-m^2)^2}$ (since $l^2+m^2=1$).
Also,$h^2 - k^2 = \frac{l^2-m^2}{(l^2-m^2)^2} = \frac{1}{l^2-m^2}$.
Thus,$h^2 + k^2 = (h^2 - k^2)^2$.
The locus is $x^2 + y^2 = (x^2 - y^2)^2$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
For $\Delta PQR$ with $P(x, y), Q(a, 2a), R(-a, -2a)$:
Area $= \frac{1}{2} |x(2a - (-2a)) + a(-2a - y) + (-a)(y - 2a)| = \frac{1}{2} |4ax - 2a^2 - ay - ay + 2a^2| = \frac{1}{2} |4ax - 2ay| = |2ax - ay|$.
For $\Delta PST$ with $P(x, y), S(a, 2a), T(2a, 3a)$:
Area $= \frac{1}{2} |x(2a - 3a) + a(3a - y) + 2a(y - 2a)| = \frac{1}{2} |-ax + 3a^2 - ay + 2ay - 4a^2| = \frac{1}{2} |-ax + ay - a^2| = \frac{1}{2} |ay - ax - a^2|$.
Equating the areas: $|2ax - ay| = \frac{1}{2} |ay - ax - a^2|$.
Case $1$: $2ax - ay = \frac{1}{2}(ay - ax - a^2) \implies 4ax - 2ay = ay - ax - a^2 \implies 5ax - 3ay = -a^2 \implies 5x - 3y = -a$.
Case $2$: $2ax - ay = -\frac{1}{2}(ay - ax - a^2) \implies 4ax - 2ay = -ay + ax + a^2 \implies 3ax - ay = a^2 \implies 3x - y = a$ (assuming $a \neq 0$).
Comparing with options,$3x - y = a$ is the correct locus.

(B) Let the square have vertices at $(\pm 1, \pm 1)$. The area of the square is $2 \times 2 = 4$ square units.
The intersection of the diagonals is the origin $O(0, 0)$.
$A$ point $P(x, y)$ is nearer to $O$ than to a vertex $V(1, 1)$ if $PO < PV$,which implies $PO^2 < PV^2$.
$x^2 + y^2 < (x-1)^2 + (y-1)^2$
$x^2 + y^2 < x^2 - 2x + 1 + y^2 - 2y + 1$
$0 < -2x - 2y + 2 \implies x + y < 1$.
Similarly,considering all four vertices,the region is defined by the inequalities $|x| + |y| < 1$.
This region is a square with vertices at $(1, 0), (0, 1), (-1, 0), (0, -1)$.
The side length of this inner square is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
The area of this region is $(\sqrt{2})^2 = 2$ square units.

(B) Let the point $P$ be $(x, y)$. The perpendicular distances from $P$ to the lines $L_1: 2x + y - 3 = 0$ and $L_2: x - 2y + 1 = 0$ are given by $d_1 = \frac{|2x + y - 3|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y - 3|}{\sqrt{5}}$ and $d_2 = \frac{|x - 2y + 1|}{\sqrt{1^2 + (-2)^2}} = \frac{|x - 2y + 1|}{\sqrt{5}}$.
Given $d_1 + d_2 = 2$,we have $\frac{|2x + y - 3| + |x - 2y + 1|}{\sqrt{5}} = 2$,which simplifies to $|2x + y - 3| + |x - 2y + 1| = 2\sqrt{5}$.
The locus of $P$ forms a square. The distance between the parallel lines formed by the absolute values is $2\sqrt{5}$.
The side length $s$ of the square is given by $s = \frac{2\sqrt{5}}{\sqrt{2^2+1^2}} \times \sqrt{2} = 2\sqrt{2}$.
Alternatively,the area of the square formed by $|ax+by+c_1| + |bx-ay+c_2| = k$ is $\frac{2k^2}{a^2+b^2}$.
Here,$a=2, b=1, k=2\sqrt{5}$.
Area $= \frac{2(2\sqrt{5})^2}{2^2+1^2} = \frac{2(20)}{5} = \frac{40}{5} = 8 \, sq \, units$.

(B) Let the first term of the $AP$ be $p$ and the common difference be $q$.
Then,the terms are given by $a = p + q$,$b = p + 3q$,and $c = p + 6q$.
The equation of the line is $ax + by + c = 0$.
Substituting the values of $a, b, c$:
$(p + q)x + (p + 3q)y + (p + 6q) = 0$
Rearranging the terms to group $p$ and $q$:
$p(x + y + 1) + q(x + 3y + 6) = 0$
For this line to pass through a fixed point for all $p$ and $q$,both expressions must be zero:
$x + y + 1 = 0$ (Equation $1$)
$x + 3y + 6 = 0$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(x + 3y + 6) - (x + y + 1) = 0$
$2y + 5 = 0 \Rightarrow y = -\frac{5}{2}$
Substituting $y = -\frac{5}{2}$ into Equation $1$:
$x - \frac{5}{2} + 1 = 0$ $\Rightarrow x - \frac{3}{2} = 0$ $\Rightarrow x = \frac{3}{2}$
Thus,the fixed point is $\left( \frac{3}{2}, -\frac{5}{2} \right)$.

(B) Let the coordinates of $A$ be $(1, 2)$.
The line containing $B$ and $C$ is $L: x - y + \lambda = 0$.
The altitude from $A$ to the line $BC$ is perpendicular to $BC$.
The slope of $BC$ is $m = 1$.
Therefore,the slope of the altitude from $A$ is $m' = -1$.
The equation of the altitude from $A$ is $y - 2 = -1(x - 1)$,which simplifies to $y - 2 = -x + 1$,or $x + y = 3$.
Since the orthocenter of a triangle always lies on its altitudes,and the altitude from $A$ is fixed regardless of the value of $\lambda$,the locus of the orthocenter is the line $x + y = 3$.

(D) The line is $y = \sqrt{3}x$. This can be written in parametric form as $x = r \cos(\theta)$ and $y = r \sin(\theta)$,where $\tan(\theta) = \sqrt{3}$,so $\theta = 60^\circ$. Thus,$x = r \cos(60^\circ) = \frac{r}{2}$ and $y = r \sin(60^\circ) = \frac{r\sqrt{3}}{2}$.
Substituting these into the curve equation $x^3 + 3y^2 + 4x + 5y - 1 = 0$:
$(\frac{r}{2})^3 + 3(\frac{r\sqrt{3}}{2})^2 + 4(\frac{r}{2}) + 5(\frac{r\sqrt{3}}{2}) - 1 = 0$
$\frac{r^3}{8} + \frac{9r^2}{4} + 2r + \frac{5\sqrt{3}r}{2} - 1 = 0$
Multiplying by $8$:
$r^3 + 18r^2 + (16 + 20\sqrt{3})r - 8 = 0$
Let the roots be $r_1, r_2, r_3$,which represent the distances $OA, OB, OC$. The product of the roots is given by the constant term with a sign change (for a cubic $ar^3 + br^2 + cr + d = 0$,product is $-d/a$):
$OA \cdot OB \cdot OC = r_1 r_2 r_3 = -(-8)/1 = 8$.

(B) Let the foot of the perpendicular from the origin $(0,0)$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(h, k).$
Since $P(h, k)$ lies on the line,we have $\frac{h}{a} + \frac{k}{b} = 1.$
Also,the slope of the line $OP$ is $\frac{k}{h}$ and the slope of the given line is $-\frac{b}{a}.$
Since $OP$ is perpendicular to the line,their product of slopes is $-1$: $\left(\frac{k}{h}\right) \left(-\frac{b}{a}\right) = -1$ $\Rightarrow \frac{b}{a} = \frac{h}{k}$ $\Rightarrow a = \frac{bh}{k}.$
Substituting $a$ into the line equation: $\frac{hk}{bh} + \frac{y}{b} = 1$ $\Rightarrow \frac{k}{b} + \frac{y}{b} = 1$ $\Rightarrow b = k + \frac{ky}{h}$ (This is not correct,let's use the standard approach).
Alternative approach: The line is $\frac{x}{a} + \frac{y}{b} = 1.$ The foot of the perpendicular $(h, k)$ satisfies $h = \frac{a b^2}{a^2 + b^2}$ and $k = \frac{a^2 b}{a^2 + b^2}.$
Then $h^2 + k^2 = \frac{a^2 b^2 (b^2 + a^2)}{(a^2 + b^2)^2} = \frac{a^2 b^2}{a^2 + b^2} = \frac{1}{\frac{1}{b^2} + \frac{1}{a^2}}.$
Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ we have $h^2 + k^2 = \frac{1}{1/(2c^2)} = 2c^2.$
Thus,the locus is $x^2 + y^2 = 2c^2.$

(B) Let the equation of the variable line passing through $(a, b)$ be $\frac{x}{X} + \frac{y}{Y} = 1$.
Since it passes through $(a, b)$,we have $\frac{a}{X} + \frac{b}{Y} = 1$.
The line meets the axes at $A(X, 0)$ and $B(0, Y)$.
The lines drawn through $A$ and $B$ parallel to the coordinate axes intersect at point $P(h, k)$.
From the geometry,$h = X$ and $k = Y$.
Substituting these into the equation $\frac{a}{X} + \frac{b}{Y} = 1$,we get $\frac{a}{h} + \frac{b}{k} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{a}{x} + \frac{b}{y} = 1$.

(C) Let $P = (x, y)$.
Given $A = (2, 3)$ and $B = (-4, 5)$.
The area of $\Delta PAB = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$.
Substituting the coordinates: $\frac{1}{2} |x(3 - 5) + 2(5 - y) - 4(y - 3)| = 12$.
$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$.
$|-2x - 6y + 22| = 24$.
$|-x - 3y + 11| = 12$.
This implies $-x - 3y + 11 = 12$ or $-x - 3y + 11 = -12$.
Case $1$: $x + 3y + 1 = 0$.
Case $2$: $x + 3y - 23 = 0$.
The combined equation of the locus is $(x + 3y + 1)(x + 3y - 23) = 0$.
Expanding this: $x^2 + 3xy - 23x + 3xy + 9y^2 - 69y + x + 3y - 23 = 0$.
$x^2 + 6xy + 9y^2 - 22x - 66y - 23 = 0$.

(A) Let the two mutually perpendicular lines be the coordinate axes $x=0$ and $y=0$.
Let the point be $(x, y)$.
The sum of its distances from these lines is $|x| + |y| = 3$.
This equation represents a square with vertices at $(3, 0), (0, 3), (-3, 0),$ and $(0, -3)$.
The diagonals of this square lie along the axes and have lengths $d_1 = 6$ and $d_2 = 6$.
The area of the square is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 6 \times 6 = 18 \text{ unit}^2$.

(D) Let the line $AB$ have slope $m$. Since $AB$ passes through $A(1, 1)$,its equation is $y - 1 = m(x - 1)$.
It intersects the $x$-axis at $B(1 - \frac{1}{m}, 0)$.
The line $AC$ is perpendicular to $AB$,so its slope is $-\frac{1}{m}$. Its equation is $y - 1 = -\frac{1}{m}(x - 1)$.
It intersects the $y$-axis at $C(0, 1 + m)$.
Let the centroid of $\Delta ABC$ be $G(h, k)$.
$h = \frac{1 + (1 - \frac{1}{m}) + 0}{3} = \frac{2 - \frac{1}{m}}{3} \implies 3h - 2 = -\frac{1}{m} \implies m = \frac{1}{2 - 3h}$.
$k = \frac{1 + 0 + (1 + m)}{3} = \frac{2 + m}{3} \implies 3k - 2 = m$.
Substituting $m$ in the second equation: $3k - 2 = \frac{1}{2 - 3h}$.
$(3k - 2)(2 - 3h) = 1 \implies 6k - 9hk - 4 + 6h = 1$.
$9hk - 6h - 6k + 5 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $9xy - 6x - 6y + 5 = 0$.

(A) The expression $|PR + QR| + |PR - QR|$ represents the sum of the distances from $R$ to $P$ and $Q$ plus the absolute difference of those distances.
By the triangle inequality,$|PR + QR| \ge |PQ|$.
Also,$|PR - QR| \ge 0$,and the minimum value is $0$ when $R$ lies on the line segment $PQ$.
Thus,the minimum value of the entire expression is $|PQ|$,which occurs when $R$ lies on the line segment $PQ$.
However,for the expression to be minimized specifically as a function of $R$,$R$ must lie on the line segment $PQ$.
Given the standard interpretation of such geometric minimization problems,$R$ is the midpoint of $PQ$.
$R = \left( \frac{2+6}{2}, \frac{3+0}{2} \right) = (4, 1.5)$.
So,$\alpha = 4$ and $\beta = 1.5$.
Then,$\alpha - 2\beta = 4 - 2(1.5) = 4 - 3 = 1$.

(C) The given lines are $L_1: 4x + 3y - 12 = 0$ and $L_2: 3x + 4y - 12 = 0$.
Any line passing through their intersection is given by $L_1 + \lambda L_2 = 0$,which is $(4x + 3y - 12) + \lambda(3x + 4y - 12) = 0$.
Rearranging,we get $x(4 + 3\lambda) + y(3 + 4\lambda) - 12(1 + \lambda) = 0$.
The line meets the axes at $A\left(\frac{12(1 + \lambda)}{4 + 3\lambda}, 0\right)$ and $B\left(0, \frac{12(1 + \lambda)}{3 + 4\lambda}\right)$.
Let the midpoint of $AB$ be $(h, k)$. Then $h = \frac{6(1 + \lambda)}{4 + 3\lambda}$ and $k = \frac{6(1 + \lambda)}{3 + 4\lambda}$.
From these,$\frac{1}{h} = \frac{4 + 3\lambda}{6(1 + \lambda)}$ and $\frac{1}{k} = \frac{3 + 4\lambda}{6(1 + \lambda)}$.
Adding these,$\frac{1}{h} + \frac{1}{k} = \frac{7 + 7\lambda}{6(1 + \lambda)} = \frac{7}{6}$.
Thus,$\frac{h + k}{hk} = \frac{7}{6}$,which implies $6(h + k) = 7hk$.
The locus is $6(x + y) = 7xy$.

(D) Let the abscissa of $Q$ be $x$.
Since $Q$ lies on the $x$-axis,its coordinates are $Q(x, 0)$.
Let the reflected ray make an angle $\theta$ with the positive $x$-axis.
The slope of the reflected ray $QR$ is $\tan \theta = \frac{7 - 0}{6 - x} = \frac{7}{6 - x}$.
The incident ray $PQ$ makes an angle $(180^\circ - \theta)$ with the positive $x$-axis.
The slope of the incident ray $PQ$ is $\tan(180^\circ - \theta) = \frac{3 - 0}{1 - x} = \frac{3}{1 - x}$.
Since $\tan(180^\circ - \theta) = -\tan \theta$,we have:
$\frac{3}{1 - x} = -\left(\frac{7}{6 - x}\right)$
$3(6 - x) = -7(1 - x)$
$18 - 3x = -7 + 7x$
$25 = 10x$
$x = \frac{25}{10} = \frac{5}{2}$.

(C) Let the coordinates of point $P$ be $(\alpha, \beta)$. Since $P$ lies on the line $2x - 3y + 4 = 0$,we have $2\alpha - 3\beta + 4 = 0$.
Let $(h, k)$ be the centroid of $\Delta PQR$. The coordinates of the centroid are given by:
$h = \frac{\alpha + 1 + 3}{3}$ $\Rightarrow 3h = \alpha + 4$ $\Rightarrow \alpha = 3h - 4$
$k = \frac{\beta + 4 - 2}{3}$ $\Rightarrow 3k = \beta + 2$ $\Rightarrow \beta = 3k - 2$
Substituting these values into the equation of the line:
$2(3h - 4) - 3(3k - 2) + 4 = 0$
$6h - 8 - 9k + 6 + 4 = 0$
$6h - 9k + 2 = 0$
The locus of the centroid $(h, k)$ is $6x - 9y + 2 = 0$.
Rewriting in slope-intercept form: $9y = 6x + 2$ $\Rightarrow y = \frac{6}{9}x + \frac{2}{9}$ $\Rightarrow y = \frac{2}{3}x + \frac{2}{9}$.
The slope of this line is $\frac{2}{3}$.

(D) Let the point be $P(t, y)$. Since the point lies on the line $3x + 5y = 15$,we have $3t + 5y = 15$,which gives $y = \frac{15 - 3t}{5}$.
Since the point is equidistant from the coordinate axes,$|x| = |y|$,so $|t| = |\frac{15 - 3t}{5}|$.
This implies $\frac{15 - 3t}{5} = t$ or $\frac{15 - 3t}{5} = -t$.
Case $1$: $15 - 3t = 5t$ $\Rightarrow 8t = 15$ $\Rightarrow t = \frac{15}{8}$. Then $y = \frac{15}{8}$. Point $P(\frac{15}{8}, \frac{15}{8})$ lies in the $1^{st}$ quadrant.
Case $2$: $15 - 3t = -5t$ $\Rightarrow 2t = -15$ $\Rightarrow t = -\frac{15}{2}$. Then $y = \frac{15}{2}$. Point $P(-\frac{15}{2}, \frac{15}{2})$ lies in the $2^{nd}$ quadrant.
Thus,the points lie in the $1^{st}$ and $2^{nd}$ quadrants.

(C) Let a point on the line $x=2y$ be $P(2\alpha, \alpha)$.
Let the perpendicular from $P$ to the line $x-y=0$ meet it at $Q(\beta, \beta)$.
The slope of $PQ$ is $\frac{\alpha-\beta}{2\alpha-\beta}$.
Since $PQ$ is perpendicular to $x-y=0$ (slope $1$),the slope of $PQ$ must be $-1$.
So,$\frac{\alpha-\beta}{2\alpha-\beta} = -1 \implies \alpha-\beta = -2\alpha+\beta \implies 3\alpha = 2\beta \implies \beta = \frac{3\alpha}{2}$.
Let $(h, k)$ be the mid-point of $PQ$.
$h = \frac{2\alpha+\beta}{2} = \frac{2\alpha + \frac{3\alpha}{2}}{2} = \frac{7\alpha}{4}$.
$k = \frac{\alpha+\beta}{2} = \frac{\alpha + \frac{3\alpha}{2}}{2} = \frac{5\alpha}{4}$.
Now,$\frac{h}{k} = \frac{7\alpha/4}{5\alpha/4} = \frac{7}{5}$.
$5h = 7k \implies 5x-7y=0$.

(N/A) Given lines are $3x - 2y - 5 = 0$ $(1)$ and $3x + 2y - 5 = 0$ $(2)$.
Let $(h, k)$ be any point whose distances from lines $(1)$ and $(2)$ are equal.
The distance of $(h, k)$ from line $(1)$ is $\frac{|3h - 2k - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3h - 2k - 5|}{\sqrt{13}}$.
The distance of $(h, k)$ from line $(2)$ is $\frac{|3h + 2k - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3h + 2k - 5|}{\sqrt{13}}$.
Since the distances are equal,we have $\frac{|3h - 2k - 5|}{\sqrt{13}} = \frac{|3h + 2k - 5|}{\sqrt{13}}$,which implies $|3h - 2k - 5| = |3h + 2k - 5|$.
This gives two cases:
Case $1$: $3h - 2k - 5 = 3h + 2k - 5 \implies -2k = 2k \implies 4k = 0 \implies k = 0$.
Case $2$: $3h - 2k - 5 = -(3h + 2k - 5) \implies 3h - 2k - 5 = -3h - 2k + 5 \implies 6h = 10 \implies h = \frac{5}{3}$.
Thus,the locus of the point $(h, k)$ is $y = 0$ or $x = \frac{5}{3}$,both of which represent straight lines.

(N/A) The equations of the given lines are:
$x + y - 5 = 0$ $(1)$
$3x - 2y + 7 = 0$ $(2)$
The perpendicular distances of $P(x, y)$ from lines $(1)$ and $(2)$ are given by:
$d_{1} = \frac{|x + y - 5|}{\sqrt{1^{2} + 1^{2}}} = \frac{|x + y - 5|}{\sqrt{2}}$
$d_{2} = \frac{|3x - 2y + 7|}{\sqrt{3^{2} + (-2)^{2}}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$
Given that $d_{1} + d_{2} = 10$,we have:
$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$
Assuming the expressions inside the modulus are positive,we get:
$\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$
Expanding the terms:
$(\sqrt{13} + 3\sqrt{2})x + (\sqrt{13} - 2\sqrt{2})y - (5\sqrt{13} - 7\sqrt{2} + 10\sqrt{26}) = 0$
This is of the form $Ax + By + C = 0$,which represents a straight line. Since the signs of the expressions inside the modulus can change,$P$ moves on one of the four possible lines formed by the combinations of signs. Thus,$P$ moves on a line.

(A) Let the coordinates of point $A$ be $(a, 0)$.
By the law of reflection,the point $(1, 2)$ and the point $(5, 3)$ are such that the reflection of $(1, 2)$ across the $x$-axis,which is $(1, -2)$,lies on the line passing through $A$ and $(5, 3)$.
The slope of the line passing through $(1, -2)$ and $(5, 3)$ is $m = \frac{3 - (-2)}{5 - 1} = \frac{5}{4}$.
The equation of this line is $y - 3 = \frac{5}{4}(x - 5)$.
To find the $x$-intercept (point $A$),set $y = 0$:
$-3 = \frac{5}{4}(x - 5)$
$-12 = 5x - 25$
$5x = 13$
$x = \frac{13}{5}$.
Thus,the coordinates of point $A$ are $\left(\frac{13}{5}, 0\right)$.

(A) Let the pencil be rotated by an angle $\alpha$ about $C$. Let the new position of the pencil be $A'B'$.
In the right-angled triangle formed by the perpendicular from $P$ to the pencil,let the angle between $PC$ and the new pencil line be $\theta$.
We are given $PC = \sqrt{5}$ and the perpendicular distance $d = 1$.
In the right triangle formed by $P$,the foot of the perpendicular on the pencil,and $C$,we have $\sin \theta = \frac{d}{PC} = \frac{1}{\sqrt{5}}$.
Thus,$\tan \theta = \frac{1}{2}$.
Initially,the angle $\angle PCB = \phi = \tan^{-1}(2)$,so $\tan \phi = 2$.
The angle of rotation $\alpha$ is the difference between the initial angle $\phi$ and the new angle $\theta$.
$\alpha = \phi - \theta$.
$\tan \alpha = \tan(\phi - \theta) = \frac{\tan \phi - \tan \theta}{1 + \tan \phi \tan \theta} = \frac{2 - 1/2}{1 + 2(1/2)} = \frac{3/2}{2} = \frac{3}{4}$.
Therefore,$\alpha = \tan^{-1}\left(\frac{3}{4}\right)$.

(C) Given $A = (0,6)$ and $B = (2t, 0)$.
The mid-point $M$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{6+0}{2}\right) = (t, 3)$.
The slope of $AB$ is $m_{AB} = \frac{0-6}{2t-0} = \frac{-6}{2t} = -\frac{3}{t}$.
The slope of the perpendicular bisector of $AB$ is $m_{\perp} = -\frac{1}{m_{AB}} = \frac{t}{3}$.
The equation of the perpendicular bisector passing through $M(t, 3)$ is $(y-3) = \frac{t}{3}(x-t)$.
To find $C$,we set $x=0$ in the equation: $y-3 = \frac{t}{3}(0-t) \Rightarrow y = 3 - \frac{t^{2}}{3}$.
So,$C = (0, 3 - \frac{t^{2}}{3})$.
Let $P(h, k)$ be the mid-point of $MC$. Since $M = (t, 3)$ and $C = (0, 3 - \frac{t^{2}}{3})$:
$h = \frac{t+0}{2} = \frac{t}{2} \Rightarrow t = 2h$.
$k = \frac{3 + (3 - \frac{t^{2}}{3})}{2} = \frac{6 - \frac{t^{2}}{3}}{2} = 3 - \frac{t^{2}}{6}$.
Substituting $t = 2h$ into the equation for $k$:
$k = 3 - \frac{(2h)^{2}}{6} = 3 - \frac{4h^{2}}{6} = 3 - \frac{2h^{2}}{3}$.
$3k = 9 - 2h^{2} \Rightarrow 2h^{2} + 3k - 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2x^{2} + 3y - 9 = 0$.

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12.$
Substituting the given points $(5, 6), (3, 2),$ and $(\alpha, \beta)$:
$\frac{1}{2} |5(2 - \beta) + 3(\beta - 6) + \alpha(6 - 2)| = 12$
$|10 - 5\beta + 3\beta - 18 + 4\alpha| = 24$
$|4\alpha - 2\beta - 8| = 24$
$|2\alpha - \beta - 4| = 12$
This gives two possible lines for the locus of point $A$:
Case $1$: $2\alpha - \beta - 4 = 12 \Rightarrow 2\alpha - \beta - 16 = 0$
Case $2$: $2\alpha - \beta - 4 = -12 \Rightarrow 2\alpha - \beta + 8 = 0$
The distance from the origin $(0, 0)$ to a line $ax + by + c = 0$ is $\frac{|c|}{\sqrt{a^2 + b^2}}.$
For line $1$: $d_1 = \frac{|-16|}{\sqrt{2^2 + (-1)^2}} = \frac{16}{\sqrt{5}}$
For line $2$: $d_2 = \frac{|8|}{\sqrt{2^2 + (-1)^2}} = \frac{8}{\sqrt{5}}$
The least distance is $\frac{8}{\sqrt{5}}.$

(D) To minimize the time taken at a constant speed,the total distance $PR + RQ$ must be minimized.
Let $Q'(0, -2)$ be the reflection of $Q(0, 2)$ across the $x$-axis.
The distance $RQ = RQ'$. Thus,$PR + RQ = PR + RQ'$.
This sum is minimized when $P, R,$ and $Q'$ are collinear.
The line passing through $P(-3, 4)$ and $Q'(0, -2)$ has the equation:
$y - (-2) = \frac{4 - (-2)}{-3 - 0}(x - 0)$
$y + 2 = \frac{6}{-3}x$
$y + 2 = -2x \implies 2x + y + 2 = 0$.
The point $R$ is the intersection of this line with the $x$-axis $(y=0)$:
$2x + 0 + 2 = 0 \implies x = -1$.
So,$R = (-1, 0)$.
Now,calculate the squared distances:
$PR^{2} = (-1 - (-3))^{2} + (0 - 4)^{2} = (2)^{2} + (-4)^{2} = 4 + 16 = 20$.
$RQ^{2} = (0 - (-1))^{2} + (2 - 0)^{2} = (1)^{2} + (2)^{2} = 1 + 4 = 5$.
Finally,calculate $50(PR^{2} + RQ^{2})$:
$50(20 + 5) = 50(25) = 1250$.

(D) For $\triangle ABC$ to be isosceles,at least two sides must be equal. Let the fixed segment be $BC$ with length $a$.
Case $I$: $AB = AC$. The locus of $A$ is the perpendicular bisector of the segment $BC$,which is a straight line.
Case $II$: $AB = BC$. Since $BC$ is fixed,$AB$ must be equal to the constant length $BC$. Thus,the locus of $A$ is a circle centered at $B$ with radius equal to $BC$.
Case $III$: $AC = BC$. Similarly,the locus of $A$ is a circle centered at $C$ with radius equal to $BC$.
Therefore,the complete locus of $A$ is the union of the perpendicular bisector of $BC$ and two circles centered at $B$ and $C$ with radius $BC$ (excluding points where $A, B, C$ are collinear,which are the finitely many exceptional points).
Thus,the correct option is $(d)$.

(B) regular polygon with $n$ sides can tile the plane if and only if its interior angle is a divisor of $360^{\circ}$.
The interior angle of a regular $n$-gon is given by $\frac{(n-2) \times 180^{\circ}}{n}$.
$(1)$ For an equilateral triangle $(n=3)$: Interior angle = $\frac{(3-2) \times 180^{\circ}}{3} = 60^{\circ}$. Since $360^{\circ} / 60^{\circ} = 6$,it can tile the plane.
$(2)$ For a square $(n=4)$: Interior angle = $\frac{(4-2) \times 180^{\circ}}{4} = 90^{\circ}$. Since $360^{\circ} / 90^{\circ} = 4$,it can tile the plane.
$(3)$ For a regular pentagon $(n=5)$: Interior angle = $\frac{(5-2) \times 180^{\circ}}{5} = 108^{\circ}$. Since $360^{\circ} / 108^{\circ} = 3.33$ (not an integer),it cannot tile the plane.
$(4)$ For a regular hexagon $(n=6)$: Interior angle = $\frac{(6-2) \times 180^{\circ}}{6} = 120^{\circ}$. Since $360^{\circ} / 120^{\circ} = 3$,it can tile the plane.
Thus,the shapes that can tile the plane are the equilateral triangle,the square,and the regular hexagon. There are exactly three such shapes.

(C) Let the position of the $k$-th ant at time $t$ be $x_k(t)$.
Given $x_k(0) = k^2$ and $x_k(1) = (11-k)^2$.
Since the speed is uniform,the velocity $v_k = x_k(1) - x_k(0) = (11-k)^2 - k^2 = 121 - 22k + k^2 - k^2 = 121 - 22k$.
Thus,$x_k(t) = k^2 + (121 - 22k)t$.
Two ants $i$ and $j$ (where $i < j$) are at the same location if $x_i(t) = x_j(t)$.
$i^2 + (121 - 22i)t = j^2 + (121 - 22j)t$
$i^2 - j^2 = (121 - 22j - 121 + 22i)t$
$(i-j)(i+j) = 22(i-j)t$
Since $i \neq j$,we can divide by $(i-j)$:
$t = \frac{i+j}{22}$.
Since $1 \le i < j \le 10$,the possible values for $i+j$ range from $1+2=3$ to $9+10=19$.
Thus,$t \in \{\frac{3}{22}, \frac{4}{22}, \dots, \frac{19}{22}\}$.
The number of distinct values is $19 - 3 + 1 = 17$.

(B) Let the coordinates of point $P$ be $(r, \theta)$. In Cartesian coordinates,$P = (r \cos \theta, r \sin \theta)$.
The line $OP$ passes through the origin and has the equation $y = x \tan \theta$.
The line $r \sin \theta = b$ is equivalent to $y = b$ in Cartesian coordinates.
Let $Q$ be the point of intersection of line $OP$ and the line $y = b$.
Since $Q$ lies on $y = b$,its $y$-coordinate is $b$. Since $Q$ lies on $OP$,its polar distance from the origin $OQ$ satisfies $OQ \sin \theta = b$,so $OQ = \frac{b}{\sin \theta}$.
Given $PQ = d$,the distance $OP = OQ \pm d = \frac{b}{\sin \theta} \pm d$.
Thus,$r = \frac{b}{\sin \theta} \pm d$.
Multiplying by $\sin \theta$,we get $r \sin \theta = b \pm d \sin \theta$,which simplifies to $(r \mp d) \sin \theta = b$.
Since $d$ is a constant,the locus is $(r \pm d) \sin \theta = b$.

(C) Given points are $A(4,0)$ and $B(0,12)$.
Let the point $C$ be $(x, y)$.
The area of $\triangle ABC$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the coordinates:
$18 = \frac{1}{2} |4(12 - y) + 0(y - 0) + x(0 - 12)|$
$18 = \frac{1}{2} |48 - 4y - 12x|$
$36 = |48 - 4y - 12x|$
Dividing by $4$ inside the absolute value:
$36 = 4 |12 - y - 3x|$
$9 = |-(3x + y - 12)|$
$9 = |3x + y - 12|$
Squaring both sides:
$(3x + y - 12)^2 = 81$
Thus,the locus is $(y + 3x - 12)^2 = 81$.

(D) Let the initial length of both candles be $L$.
The rate of burning for the first candle is $\frac{L}{5}$ per hour,and for the second candle is $\frac{L}{3}$ per hour.
Let $t$ be the time in hours after which the length of the first candle is $3$ times the length of the second candle.
The remaining lengths after time $t$ are $L_1 = L - \frac{L}{5}t$ and $L_2 = L - \frac{L}{3}t$.
According to the problem,$L_1 = 3L_2$.
Substituting the expressions: $L - \frac{L}{5}t = 3(L - \frac{L}{3}t)$.
Dividing by $L$: $1 - \frac{t}{5} = 3 - t$.
Rearranging the terms: $t - \frac{t}{5} = 3 - 1$.
$\frac{4t}{5} = 2$.
$t = \frac{10}{4} = 2.5 \, hr$.
Converting to minutes: $2.5 \times 60 = 150 \, min$.

(C) Let the length of the train be $x\,m$.
The time taken by the train to pass the man is $9\,s$.
Therefore,the speed of the train is $v = \frac{x}{9}\,m/s$.
The time taken by the train to cross the platform is $21\,s$.
When crossing the platform,the total distance covered is the sum of the length of the train and the length of the platform,which is $(x + 88)\,m$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$,we have:
$x + 88 = v \times 21$
Substituting $v = \frac{x}{9}$:
$x + 88 = \frac{x}{9} \times 21$
$x + 88 = \frac{7x}{3}$
$3(x + 88) = 7x$
$3x + 264 = 7x$
$4x = 264$
$x = 66\,m$.
Thus,the length of the train is $66\,m$.

(B) The frog starts at $(0,0)$ and needs to reach $(0,1)$ with each jump covering a distance of $5$ units.
Let the jump be represented by a vector $(x, y)$ such that $x^2 + y^2 = 5^2 = 25$,where $x, y \in \mathbb{Z}$.
Possible integer pairs $(x, y)$ are $(\pm 3, \pm 4)$ or $(\pm 4, \pm 3)$ or $(\pm 5, 0)$ or $(0, \pm 5)$.
To reach $(0,1)$ in the minimum number of jumps:
$1$. Jump $1$: From $(0,0)$ to $(4,3)$ (distance $5$).
$2$. Jump $2$: From $(4,3)$ to $(0,6)$ (distance $\sqrt{(0-4)^2 + (6-3)^2} = \sqrt{16+9} = 5$).
$3$. Jump $3$: From $(0,6)$ to $(0,1)$ (distance $5$).
Thus,the minimum number of jumps required is $3$.

(B) Let the square $ABCD$ be placed in the coordinate plane with $A(0, 12), B(12, 12), C(12, 0), D(0, 0)$.
$M$ is the midpoint of $AB$,so $M = (6, 12)$.
$N$ is the midpoint of $CD$,so $N = (6, 0)$.
$P$ lies on $MN$,so $P = (6, y)$ for some $y \in [0, 12]$.
$AP = r = \sqrt{(6-0)^2 + (y-12)^2} = \sqrt{36 + (12-y)^2}$.
$PC = s = \sqrt{(6-12)^2 + (y-0)^2} = \sqrt{36 + y^2}$.
Consider the triangle with sides $r, s, 12$. Note that $BC = 12$. The point $P$ has a horizontal distance of $6$ from the side $BC$ (which lies on the line $x=12$).
Thus,the area of $\triangle PBC$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times \text{distance of } P \text{ from } BC = \frac{1}{2} \times 12 \times 6 = 36$.

(B) Let $x$ be the number of minutes past $10\,\text{O'clock}$.
At $10\,\text{O'clock}$,the hour hand is at $300^{\circ}$ from the $12\,\text{O'clock}$ position.
In $x$ minutes,the minute hand moves $6x^{\circ}$ from the $12\,\text{O'clock}$ position.
The hour hand moves $\frac{x}{2}^{\circ}$ in $x$ minutes,so its position is $(300 + \frac{x}{2})^{\circ}$ from $12\,\text{O'clock}$.
For the hands to be symmetric with respect to the vertical line (the $12-6$ line),the angle of the minute hand from $12\,\text{O'clock}$ (clockwise) must equal the angle of the hour hand from $12\,\text{O'clock}$ (counter-clockwise).
The angle of the hour hand from $12\,\text{O'clock}$ (counter-clockwise) is $360^{\circ} - (300 + \frac{x}{2})^{\circ} = (60 - \frac{x}{2})^{\circ}$.
Setting these equal: $6x = 60 - \frac{x}{2}$.
$12x = 120 - x$ $\Rightarrow 13x = 120$ $\Rightarrow x = \frac{120}{13} \approx 9.2307\,\text{minutes}$.
$0.2307 \times 60 \approx 13.84\,\text{seconds}$,which rounds to $14\,\text{seconds}$.
Thus,the time is $10\,\text{h } 9\,\text{m } 14\,\text{s}$.