Locus of Point Questions in English

(B) Let the point be $P(x, x)$ since the abscissa is equal to the ordinate.
Given that $P$ is equidistant from $A(1, 0)$ and $B(0, 3)$,we have $PA^2 = PB^2$.
$(x - 1)^2 + (x - 0)^2 = (x - 0)^2 + (x - 3)^2$
$(x - 1)^2 + x^2 = x^2 + (x - 3)^2$
$x^2 - 2x + 1 + x^2 = x^2 + x^2 - 6x + 9$
$-2x + 1 = -6x + 9$
$4x = 8$
$x = 2$
Therefore,the point is $(2, 2)$.

(D) Let the given points be $P(a + b, b - a)$ and $Q(a - b, a + b)$. The point $(x, y)$ is equidistant from $P$ and $Q$,so $dist(P, (x, y))^2 = dist(Q, (x, y))^2$.
$(x - (a + b))^2 + (y - (b - a))^2 = (x - (a - b))^2 + (y - (a + b))^2$
Expanding both sides:
$x^2 - 2x(a + b) + (a + b)^2 + y^2 - 2y(b - a) + (b - a)^2 = x^2 - 2x(a - b) + (a - b)^2 + y^2 - 2y(a + b) + (a + b)^2$
Canceling $x^2, y^2, (a + b)^2$ from both sides:
$-2ax - 2bx + (b - a)^2 - 2by + 2ay = -2ax + 2bx + (a - b)^2 - 2ay - 2by$
Since $(b - a)^2 = (a - b)^2$,these terms also cancel:
$-2bx + 2ay = 2bx - 2ay$
$4ay = 4bx$
$bx - ay = 0$
Thus,the correct option is $D$.

(D) Let the point be $P(x, y)$. The distance from $P$ to $(2, 0)$ is equal to the distance from $P$ to $(0, 2)$.
Using the distance formula: $\sqrt{(x-2)^2 + (y-0)^2} = \sqrt{(x-0)^2 + (y-2)^2}$.
Squaring both sides: $(x-2)^2 + y^2 = x^2 + (y-2)^2$.
Expanding the terms: $x^2 - 4x + 4 + y^2 = x^2 + y^2 - 4y + 4$.
Simplifying: $-4x = -4y$,which implies $x = y$.
Checking the options,the point $(2, 2)$ satisfies $x = y$ and is equidistant from $(2, 0)$ and $(0, 2)$.

(A) Since points $A(3, 4)$,$B(7, 7)$,and $C(a, b)$ are collinear,the slope of $AB$ must equal the slope of $BC$.
Slope of $AB = \frac{7 - 4}{7 - 3} = \frac{3}{4}$.
Slope of $BC = \frac{b - 7}{a - 7} = \frac{3}{4}$,which implies $4(b - 7) = 3(a - 7)$,or $4b - 28 = 3a - 21$,so $3a - 4b = -7$.
Given $AC = 10$,we have $\sqrt{(a - 3)^2 + (b - 4)^2} = 10$,so $(a - 3)^2 + (b - 4)^2 = 100$.
Substituting $b = \frac{3a + 7}{4}$ into the distance equation:
$(a - 3)^2 + (\frac{3a + 7}{4} - 4)^2 = 100$
$(a - 3)^2 + (\frac{3a - 9}{4})^2 = 100$
$(a - 3)^2 + \frac{9(a - 3)^2}{16} = 100$
$\frac{25(a - 3)^2}{16} = 100 \implies (a - 3)^2 = 64 \implies a - 3 = \pm 8$.
If $a = 11$,$b = \frac{3(11) + 7}{4} = 10$. If $a = -5$,$b = \frac{3(-5) + 7}{4} = -2$.
Comparing with options,the correct point is $(11, 10)$.

(D) Let the point $P$ be $(x, y)$.
Since $O$ is the origin $(0, 0)$,the slope of the segment $OP$ is given by $m = \frac{y - 0}{x - 0} = \frac{y}{x}$.
Given that the slope is $\sqrt{3}$,we have $\frac{y}{x} = \sqrt{3}$.
Multiplying both sides by $x$,we get $y = \sqrt{3}x$.
Rearranging the terms,we obtain $\sqrt{3}x - y = 0$.
Thus,the required locus is $\sqrt{3}x - y = 0$.

(A) Let the coordinates of the moving point $P$ be $(x, y)$.
The coordinates of the given points are $O(0, 0)$,$A(0, 4)$,and $B(6, 0)$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area of $\Delta POA = \frac{1}{2} |0(4 - y) + 0(y - 0) + x(0 - 4)| = \frac{1}{2} |-4x| = 2|x|$.
Area of $\Delta POB = \frac{1}{2} |0(0 - y) + 6(y - 0) + x(0 - 0)| = \frac{1}{2} |6y| = 3|y|$.
According to the problem,Area of $\Delta POA = 2 \times$ Area of $\Delta POB$.
Therefore,$2|x| = 2 \times 3|y|$.
$|x| = 3|y|$.
Squaring both sides,we get $x^2 = 9y^2$,which implies $x^2 - 9y^2 = 0$.
This can be factored as $(x - 3y)(x + 3y) = 0$.

(A) The origin $O$ is $(0,0)$ and point $A$ is $(3,4)$.
Since the line segment $OP$ is parallel to $OA$ and both share the common point $O$,the points $O, A,$ and $P$ must be collinear.
The slope of line $OA$ is $m = \frac{4-0}{3-0} = \frac{4}{3}$.
Since $P(x,y)$ lies on the line passing through $O(0,0)$ with slope $m = \frac{4}{3}$,the equation of the locus is $y - 0 = \frac{4}{3}(x - 0)$.
This simplifies to $3y = 4x$,or $4x - 3y = 0$.

(B) Let the moving point be $P(h, k)$.
Given that the distance $PA = PB$.
Using the distance formula,we have:
$\sqrt{(h - a)^2 + (k - 0)^2} = \sqrt{(h - (-a))^2 + (k - 0)^2}$
Squaring both sides:
$(h - a)^2 + k^2 = (h + a)^2 + k^2$
$h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2 + k^2$
$-2ah = 2ah$
$4ah = 0$
Since $a \neq 0$,we get $h = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x = 0$,which is the $y$-axis.
The $y$-axis is the perpendicular bisector of the line segment joining $A(a, 0)$ and $B(-a, 0)$.

(B) Let the coordinates of point $P$ be $(x, y)$.
Given $PA^2 - PB^2 = 2k^2$.
Using the distance formula,$PA^2 = (x - a)^2 + (y - 0)^2$ and $PB^2 = (x + a)^2 + (y - 0)^2$.
Substituting these into the given equation:
$((x - a)^2 + y^2) - ((x + a)^2 + y^2) = 2k^2$
$(x^2 - 2ax + a^2 + y^2) - (x^2 + 2ax + a^2 + y^2) = 2k^2$
$-4ax = 2k^2$
$4ax + 2k^2 = 0$
Dividing by $2$,we get $2ax + k^2 = 0$.

(B) The given line is $x \cos \alpha + y \sin \alpha = p$.
To find the $x$-intercept,set $y = 0$: $x \cos \alpha = p \Rightarrow x = \frac{p}{\cos \alpha}$. So,point $A = (\frac{p}{\cos \alpha}, 0)$.
To find the $y$-intercept,set $x = 0$: $y \sin \alpha = p \Rightarrow y = \frac{p}{\sin \alpha}$. So,point $B = (0, \frac{p}{\sin \alpha})$.
Let $(h, k)$ be the mid-point of the segment $AB$.
Then $h = \frac{p}{2 \cos \alpha}$ and $k = \frac{p}{2 \sin \alpha}$.
This implies $\cos \alpha = \frac{p}{2h}$ and $\sin \alpha = \frac{p}{2k}$.
Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we get $(\frac{p}{2k})^2 + (\frac{p}{2h})^2 = 1$.
$\frac{p^2}{4k^2} + \frac{p^2}{4h^2} = 1 \Rightarrow \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

(B) Let $P$ be $(x, y)$. The area of $\Delta PAB$ 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 coordinates $P(x, y)$,$A(2, 3)$,and $B(-4, 5)$:
$\frac{1}{2} |x(3 - 5) + 2(5 - y) + (-4)(y - 3)| = 12$
$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$
$|-2x - 6y + 22| = 24$
$|-x - 3y + 11| = 12$
This implies $-x - 3y + 11 = 12$ or $-x - 3y + 11 = -12$.
Case $1$: $x + 3y + 1 = 0$.
Case $2$: $x + 3y - 23 = 0$.
Therefore,the locus is $(x + 3y + 1)(x + 3y - 23) = 0$.

(B) Let the centroid of the triangle be $(x, y)$.
Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3, y_3) = (1, 0)$.
The coordinates of the centroid are given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.
Thus,$3x = a \cos t + b \sin t + 1$ and $3y = a \sin t - b \cos t$.
Rearranging,we get $3x - 1 = a \cos t + b \sin t$ and $3y = a \sin t - b \cos t$.
Squaring and adding both equations:
$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$.
Expanding the right side:
$= a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$.
$= a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t) = a^2 + b^2$.
Therefore,the locus is $(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

(A) Let the coordinates of point $P$ be $(x, y)$.
Given that $PA = PB$,we have $(PA)^2 = (PB)^2$.
Using the distance formula: $(x - (a + b))^2 + (y - (a - b))^2 = (x - (a - b))^2 + (y - (a + b))^2$.
Expanding both sides:
$(x^2 - 2x(a + b) + (a + b)^2) + (y^2 - 2y(a - b) + (a - b)^2) = (x^2 - 2x(a - b) + (a - b)^2) + (y^2 - 2y(a + b) + (a + b)^2)$.
Canceling $x^2, y^2, (a + b)^2,$ and $(a - b)^2$ from both sides:
$-2x(a + b) - 2y(a - b) = -2x(a - b) - 2y(a + b)$.
Dividing by $-2$:
$x(a + b) + y(a - b) = x(a - b) + y(a + b)$.
$ax + bx + ay - by = ax - bx + ay + by$.
$bx - by = -bx + by$.
$2bx - 2by = 0$.
$x - y = 0$.

(C) Since $ABC$ is an isosceles triangle with base $BC$,vertex $A(x, y)$ must lie on the perpendicular bisector of segment $BC$.
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 $m_{BC} = \frac{7 - 3}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$.
The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{BC}} = \frac{3}{4}$.
The equation of the perpendicular bisector is $y - 5 = \frac{3}{4}(x + \frac{1}{2})$.
$4y - 20 = 3x + \frac{3}{2}$ $\Rightarrow 8y - 40 = 6x + 3$ $\Rightarrow 6x - 8y + 43 = 0$.
Checking option $(C)$: $6(\frac{5}{6}) - 8(6) + 43 = 5 - 48 + 43 = 0$. Since this point satisfies the equation,it is a valid coordinate for vertex $A$.

(C) Let the rod be $AB$ with length $l$. Let the ends $A$ and $B$ lie on the $x$-axis and $y$-axis respectively,such that $A = (a, 0)$ and $B = (0, b)$.
Since the length of the rod is $l$,we have $a^2 + b^2 = l^2$.
Let $P(h, k)$ be the point on the rod that divides it in the ratio $1 : 2$. Using the section formula,we have:
$h = \frac{1 \times 0 + 2 \times a}{1 + 2} = \frac{2a}{3} \Rightarrow a = \frac{3h}{2}$
$k = \frac{1 \times b + 2 \times 0}{1 + 2} = \frac{b}{3} \Rightarrow b = 3k$
Substituting these into the equation $a^2 + b^2 = l^2$,we get:
$(\frac{3h}{2})^2 + (3k)^2 = l^2$
$\frac{9h^2}{4} + 9k^2 = l^2$
$9h^2 + 36k^2 = 4l^2$
Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 36y^2 = 4l^2$.

(D) Let the third vertex $C$ be $(x, y)$.
The centroid $G$ of triangle $ABC$ is given by:
$G = \left( \frac{2 - 2 + x}{3}, \frac{-3 + 1 + y}{3} \right) = \left( \frac{x}{3}, \frac{y - 2}{3} \right)$.
Given that the centroid moves on the line $2x + 3y = 1$,we substitute the coordinates of $G$ into this equation:
$2\left( \frac{x}{3} \right) + 3\left( \frac{y - 2}{3} \right) = 1$.
Multiplying the entire equation by $3$,we get:
$2x + 3(y - 2) = 3$
$2x + 3y - 6 = 3$
$2x + 3y = 9$.
Thus,the locus of vertex $C$ is $2x + 3y = 9$.

(A) Let the two perpendicular lines be the coordinate axes. If $a$ and $b$ are the intercepts made by the moving line on the coordinate axes,then the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$ ..... $(i)$.
According to the problem,the sum of the reciprocals of the intercepts is constant,say $\frac{1}{a} + \frac{1}{b} = \frac{1}{k}$,where $k$ is a constant.
Multiplying by $k$,we get $\frac{k}{a} + \frac{k}{b} = 1$ ..... $(ii)$.
Comparing equation $(i)$ and equation $(ii)$,we can see that the line always passes through the fixed point $(k, k)$.

(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line be $(x, y)$.
Since the line passes through $(h, k)$ and $(x, y)$,its slope is $m = \frac{y - k}{x - h}$.
The slope of the perpendicular line segment from the origin to $(x, y)$ is $m' = \frac{y - 0}{x - 0} = \frac{y}{x}$.
Since the lines are perpendicular,the product of their slopes is $-1$:
$\left(\frac{y - k}{x - h}\right) \times \left(\frac{y}{x}\right) = -1$
$y(y - k) = -x(x - h)$
$y^2 - ky = -x^2 + hx$
$x^2 + y^2 - hx - ky = 0$
Thus,the required locus is $x^2 + y^2 - hx - ky = 0$.

(C) The equation of the given line is $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$
The slope of this line is $m_1 = -\frac{b}{a}$.
The slope of the perpendicular line passing through the origin $(0, 0)$ is $m_2 = \frac{a}{b}$.
The equation of the perpendicular line is $y - 0 = \frac{a}{b}(x - 0)$,which simplifies to $\frac{x}{b} - \frac{y}{a} = 0$ .....$(ii)$
Let $(x, y)$ be the foot of the perpendicular. Squaring and adding equations $(i)$ and $(ii)$:
$\left(\frac{x}{a} + \frac{y}{b}\right)^2 + \left(\frac{x}{b} - \frac{y}{a}\right)^2 = 1^2 + 0^2$
$x^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) + y^2\left(\frac{1}{b^2} + \frac{1}{a^2}\right) = 1$
Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,we substitute this into the equation:
$(x^2 + y^2)\left(\frac{1}{c^2}\right) = 1$
Therefore,the locus is $x^2 + y^2 = c^2$.

(B) Let the two perpendicular lines be the coordinate axes $OX$ and $OY$.
Let $P(x, y)$ be the point whose locus is required.
Since $P$ is in the first quadrant,$x > 0$ and $y > 0$.
The distance of $P(x, y)$ from the $Y$-axis $(OY)$ is $x$,and the distance from the $X$-axis $(OX)$ is $y$.
According to the problem,the sum of these distances is $2$ units:
$x + y = 2$
Since the point is in the first quadrant,the locus is the line segment $x + y = 2$ where $x > 0$ and $y > 0$.

(B) The equation of a line passing through the intersection of the lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ is given by the family of lines equation: $\left( \frac{x}{\alpha} + \frac{y}{\beta} - 1 \right) + \lambda \left( \frac{x}{\beta} + \frac{y}{\alpha} - 1 \right) = 0$.
Rearranging this,we get $x \left( \frac{1}{\alpha} + \frac{\lambda}{\beta} \right) + y \left( \frac{1}{\beta} + \frac{\lambda}{\alpha} \right) = 1 + \lambda$.
This line meets the axes at $A \left( \frac{1 + \lambda}{\frac{1}{\alpha} + \frac{\lambda}{\beta}}, 0 \right)$ and $B \left( 0, \frac{1 + \lambda}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)$.
Let $(h, k)$ be the midpoint of $AB$. Then $h = \frac{1 + \lambda}{2(\frac{1}{\alpha} + \frac{\lambda}{\beta})}$ and $k = \frac{1 + \lambda}{2(\frac{1}{\beta} + \frac{\lambda}{\alpha})}$.
From these,$\frac{1}{2h} = \frac{\frac{1}{\alpha} + \frac{\lambda}{\beta}}{1 + \lambda}$ and $\frac{1}{2k} = \frac{\frac{1}{\beta} + \frac{\lambda}{\alpha}}{1 + \lambda}$.
Adding these,$\frac{1}{2h} + \frac{1}{2k} = \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \lambda(\frac{1}{\alpha} + \frac{1}{\beta})}{1 + \lambda} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$.
Thus,$\frac{h + k}{2hk} = \frac{\alpha + \beta}{\alpha \beta}$,which simplifies to $\alpha \beta (h + k) = 2hk(\alpha + \beta)$.
The locus of $(h, k)$ is $\alpha \beta (x + y) = 2xy(\alpha + \beta)$.

(A) Let the point be $(x, y)$.
The area of the triangle with vertices $(x, y)$,$(1, 5)$,and $(3, -7)$ is given by the formula:
$\frac{1}{2} |x(5 - (-7)) + 1(-7 - y) + 3(y - 5)| = 21$
$\frac{1}{2} |x(12) - 7 - y + 3y - 15| = 21$
$\frac{1}{2} |12x + 2y - 22| = 21$
$|6x + y - 11| = 21$
This gives two possible equations:
$6x + y - 11 = 21 \implies 6x + y - 32 = 0$
$6x + y - 11 = -21 \implies 6x + y + 10 = 0$
Comparing with the given options,the locus is $6x + y - 32 = 0$.

(B) Let the equation of the line passing through $(1, 1)$ be $y - 1 = m(x - 1)$,where $m$ is the slope.
For the $x$-axis intersection $(y = 0)$: $-1 = m(x - 1)$ $\Rightarrow x - 1 = -\frac{1}{m}$ $\Rightarrow x = 1 - \frac{1}{m}$. So,$A = (1 - \frac{1}{m}, 0)$.
For the $y$-axis intersection $(x = 0)$: $y - 1 = m(-1) \Rightarrow y = 1 - m$. So,$B = (0, 1 - m)$.
Let the mid-point of $AB$ be $(h, k)$.
Then $h = \frac{1 - \frac{1}{m} + 0}{2} = \frac{m - 1}{2m}$ and $k = \frac{0 + 1 - m}{2} = \frac{1 - m}{2}$.
From $k = \frac{1 - m}{2}$,we get $m = 1 - 2k$.
Substitute $m$ into the expression for $h$: $2h = \frac{(1 - 2k) - 1}{1 - 2k} = \frac{-2k}{1 - 2k}$.
$2h(1 - 2k) = -2k$ $\Rightarrow 2h - 4hk = -2k$ $\Rightarrow 2h + 2k - 4hk = 0$.
Dividing by $2$,we get $h + k - 2hk = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x + y - 2xy = 0$.

(C) Let the centroid of $\Delta ABC$ be $G(h, k)$.
Given $A = (2, 5)$ and $B = (4, -11)$. Let $C = (x, y)$.
The coordinates of the centroid are given by $h = \frac{2 + 4 + x}{3}$ and $k = \frac{5 - 11 + y}{3}$.
From these,we get $x = 3h - 6$ and $y = 3k + 6$.
Since $C(x, y)$ lies on the line $9x + 7y + 4 = 0$,we substitute the values of $x$ and $y$:
$9(3h - 6) + 7(3k + 6) + 4 = 0$
$27h - 54 + 21k + 42 + 4 = 0$
$27h + 21k - 8 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $27x + 21y - 8 = 0$.
Dividing by $3$,we get $9x + 7y - \frac{8}{3} = 0$.
This line is parallel to $9x + 7y + 4 = 0$.

(A) Let the coordinates of point $A$ be $(a, 0)$.
Since the ray is reflected at the $x$-axis,the angle of incidence equals the angle of reflection.
Let the incident ray from $(1, 2)$ to $(a, 0)$ have slope $m_1 = \frac{2 - 0}{1 - a} = \frac{2}{1 - a}$.
The reflected ray from $(a, 0)$ to $(5, 3)$ has slope $m_2 = \frac{3 - 0}{5 - a} = \frac{3}{5 - a}$.
By the law of reflection,the angle made with the $x$-axis by the incident ray is $\pi - \theta$ and by the reflected ray is $\theta$,where $\tan \theta = m_2$ and $\tan(\pi - \theta) = m_1$.
Thus,$m_1 = -m_2$.
$\frac{2}{1 - a} = -\frac{3}{5 - a}$
$2(5 - a) = -3(1 - a)$
$10 - 2a = -3 + 3a$
$13 = 5a$
$a = \frac{13}{5}$.
Therefore,the coordinates of $A$ are $(\frac{13}{5}, 0)$.

(B) Let the fixed point $P$ be $(x_1, y_1)$. The equation of a line passing through $P$ with slope $m$ is $y - y_1 = m(x - x_1)$,which can be written as $mx - y + (y_1 - mx_1) = 0$.
The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Since we are considering the algebraic sum of perpendiculars,we use the signed distance $\frac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}$.
The sum of the perpendiculars from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is:
$\frac{2m - 0 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{0 - 2 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{m - 1 + y_1 - mx_1}{\sqrt{m^2 + 1}} = 0$
Simplifying the numerator:
$(2m + y_1 - mx_1) + (y_1 - mx_1 - 2) + (m - 1 + y_1 - mx_1) = 0$
$3m - 3mx_1 + 3y_1 - 3 = 0$
$3(m(1 - x_1) + (y_1 - 1)) = 0$
Since the line is variable,this equation must hold for all values of $m$. Thus,the coefficients of $m$ and the constant term must be zero independently:
$1 - x_1 = 0 \Rightarrow x_1 = 1$
$y_1 - 1 = 0 \Rightarrow y_1 = 1$
Therefore,the coordinates of $P$ are $(1, 1)$.

(A) Let the line passing through $(a, b)$ be $y - b = m(x - a)$,where $m$ is the slope.
This can be written as $mx - y + (b - ma) = 0$.
Let $(h, k)$ be the foot of the perpendicular from the origin $(0, 0)$ to this line.
The slope of the line joining $(0, 0)$ and $(h, k)$ is $k/h$.
Since the line is perpendicular to the given line,the product of their slopes is $-1$: $(m)(k/h) = -1 \implies m = -h/k$.
Substituting $m = -h/k$ into the line equation: $(-h/k)x - y + (b - (-h/k)a) = 0$.
Multiplying by $k$: $-hx - ky + bk + ah = 0 \implies hx + ky = ah + bk$.
Since $(h, k)$ lies on the line $mx - y + b - ma = 0$,we have $mh - k + b - ma = 0$.
Substituting $m = -h/k$: $(-h/k)h - k + b - (-h/k)a = 0$.
$-h^2 - k^2 + bk + ah = 0 \implies h^2 + k^2 - ah - bk = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - ax - by = 0$.

(C) Given: $x = \tan \theta + \sin \theta$ and $y = \tan \theta - \sin \theta$.
Step $1$: Calculate $x + y$ and $x - y$.
$x + y = 2 \tan \theta$
$x - y = 2 \sin \theta$
Step $2$: Express $\tan \theta$ and $\sin \theta$ in terms of $x$ and $y$.
$\tan \theta = \frac{x + y}{2}$
$\sin \theta = \frac{x - y}{2}$
Step $3$: Use the identity $\frac{1}{\sin^2 \theta} - \frac{1}{\tan^2 \theta} = \frac{1 - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$ is not directly helpful. Instead,use $\frac{1}{\sin^2 \theta} - \frac{1}{\tan^2 \theta} = \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1 - \cos^2 \theta}{\sin^2 \theta} = 1$.
Step $4$: Substitute the values:
$\frac{4}{(x - y)^2} - \frac{4}{(x + y)^2} = 1$
$4 \left[ \frac{(x + y)^2 - (x - y)^2}{(x - y)^2 (x + y)^2} \right] = 1$
$4 \left[ \frac{4xy}{(x^2 - y^2)^2} \right] = 1$
$16xy = (x^2 - y^2)^2$.
Thus,the locus is $(x^2 - y^2)^2 = 16xy$.

(D) If the equation of a curve $f(x, y) = 0$ remains unchanged when $x$ and $y$ are interchanged,it means $f(x, y) = f(y, x)$.
This property implies that for every point $(a, b)$ on the curve,the point $(b, a)$ is also on the curve.
The set of points $(a, b)$ and $(b, a)$ are reflections of each other across the line $y = x$.
Therefore,the curve is symmetric with respect to the line $y = x$.

(A) Let the coordinates of $R$ be $(x, 0)$.
Given $P = (1, 1)$ and $Q = (3, 2)$.
The distance $PR + RQ = \sqrt{(x - 1)^2 + (0 - 1)^2} + \sqrt{(x - 3)^2 + (0 - 2)^2}$.
$PR + RQ = \sqrt{x^2 - 2x + 2} + \sqrt{x^2 - 6x + 13}$.
For the minimum value of $PR + RQ$,we set the derivative with respect to $x$ to zero:
$\frac{d}{dx}(\sqrt{x^2 - 2x + 2} + \sqrt{x^2 - 6x + 13}) = 0$.
$\frac{x - 1}{\sqrt{x^2 - 2x + 2}} + \frac{x - 3}{\sqrt{x^2 - 6x + 13}} = 0$.
$\frac{x - 1}{\sqrt{x^2 - 2x + 2}} = -\frac{x - 3}{\sqrt{x^2 - 6x + 13}}$.
Squaring both sides:
$\frac{(x - 1)^2}{x^2 - 2x + 2} = \frac{(x - 3)^2}{x^2 - 6x + 13}$.
$(x^2 - 2x + 1)(x^2 - 6x + 13) = (x^2 - 6x + 9)(x^2 - 2x + 2)$.
Expanding both sides and simplifying leads to $3x^2 - 2x - 5 = 0$.
$(3x - 5)(x + 1) = 0$,which gives $x = \frac{5}{3}$ or $x = -1$.
Since $R$ lies between the projections of $P$ and $Q$ on the $x$-axis,$x$ must be in the interval $(1, 3)$.
Thus,$x = \frac{5}{3}$.
Therefore,the point $R$ is $\left( \frac{5}{3}, 0 \right)$.

(C) Given that $a, b, c$ are in harmonic progression,we have $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.
This implies $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$.
Substitute this into the equation of the line: $\frac{x}{a} + \frac{y}{b} + (\frac{2}{b} - \frac{1}{a}) = 0$.
Rearranging the terms,we get $\frac{1}{a}(x - 1) + \frac{1}{b}(y + 2) = 0$.
For this equation to hold for all $a$ and $b$,the coefficients must be zero.
Thus,$x - 1 = 0 \implies x = 1$ and $y + 2 = 0 \implies y = -2$.
The fixed point is $(1, -2)$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12$.
Substituting $A(2, 3), B(-4, 5)$,and $P(x, y)$:
$\frac{1}{2} |x(3 - 5) + 2(5 - y) + (-4)(y - 3)| = 12$
$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$
$|-2x - 6y + 22| = 24$
$|-x - 3y + 11| = 12$
This gives two cases:
Case $1: -x - 3y + 11 = 12 \implies x + 3y + 1 = 0$
Case $2: -x - 3y + 11 = -12 \implies x + 3y - 23 = 0$
Thus,the locus is $x + 3y + 1 = 0$ or $x + 3y - 23 = 0$.

(B) Let the coordinates of point $P$ be $(x, y)$.
Given that $P$ is equidistant from $A$ and $B$,we have $PA = PB$,which implies $(PA)^2 = (PB)^2$.
Using the distance formula: $(x - (a + b))^2 + (y - (b - a))^2 = (x - (a - b))^2 + (y - (a + b))^2$.
Expanding both sides:
$(x^2 - 2x(a + b) + (a + b)^2) + (y^2 - 2y(b - a) + (b - a)^2) = (x^2 - 2x(a - b) + (a - b)^2) + (y^2 - 2y(a + b) + (a + b)^2)$.
Canceling $x^2, y^2, (a+b)^2$ from both sides:
$-2x(a + b) - 2y(b - a) + (b - a)^2 = -2x(a - b) - 2y(a + b) + (a - b)^2$.
Since $(b - a)^2 = (a - b)^2$,these terms also cancel out:
$-2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2ay - 2by$.
Simplifying the equation:
$-2bx + 2ay = 2bx - 2ay$.
$4ay - 4bx = 0$.
Dividing by $4$,we get $ay - bx = 0$,which is equivalent to $bx - ay = 0$.

(A) Let the coordinates of the fixed points be $A (a, 0)$ and $B (-a, 0)$,and let the moving point be $C (h, k)$.
From the given figure,let $D$ be the projection of $C$ on the $x$-axis,so $D = (h, 0)$.
In $\Delta ADC$,$\cot A = \frac{AD}{CD} = \frac{a - h}{k}$.
In $\Delta BDC$,$\cot B = \frac{BD}{CD} = \frac{h - (-a)}{k} = \frac{h + a}{k}$.
Given $\cot A + \cot B = \lambda$,we have:
$\frac{a - h}{k} + \frac{h + a}{k} = \lambda$
$\frac{2a}{k} = \lambda$
Replacing $(h, k)$ with $(x, y)$,we get $\frac{2a}{y} = \lambda$,which simplifies to $y\lambda = 2a$.

(C) Let the centroid of the triangle be $(h, k)$.
Then,$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$.
This implies $3h - 1 = \cos \alpha + \sin \alpha$ and $3k - 2 = \sin \alpha - \cos \alpha$.
Squaring and adding both equations:
$(3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2$.
$(3h - 1)^2 + (3k - 2)^2 = (\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) + (\sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha) = 1 + 1 = 2$.
$9h^2 - 6h + 1 + 9k^2 - 12k + 4 = 2$.
$9(h^2 + k^2) - 6h - 12k + 3 = 0$.
Dividing by $3$,we get $3(h^2 + k^2) - 2h - 4k + 1 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $3(x^2 + y^2) - 2x - 4y + 1 = 0$.

(A) Let the two perpendicular lines be the coordinate axes $X$ and $Y$. The distance of a point $P(x, y)$ from the $Y$-axis is $|x|$ and from the $X$-axis is $|y|$.
Given that the sum of the distances is $1$,we have $|x| + |y| = 1$.
If the point is in the first quadrant,$x > 0, y > 0$,so $x + y = 1$.
If the point is in the second quadrant,$x < 0, y > 0$,so $-x + y = 1$.
If the point is in the third quadrant,$x < 0, y < 0$,so $-x - y = 1$.
If the point is in the fourth quadrant,$x > 0, y < 0$,so $x - y = 1$.
These four equations represent the sides of a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.

(D) Let $C = (h, k)$.
In $\triangle CDA$,$\tan A = \frac{k}{a-h}$.
In $\triangle CDB$,$\tan B = \frac{k}{h-(-a)} = \frac{k}{h+a}$.
Given $\angle A - \angle B = \theta$,so $\tan(A - B) = \tan \theta$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan \theta$,
$\frac{\frac{k}{a-h} - \frac{k}{h+a}}{1 + \frac{k}{a-h} \cdot \frac{k}{h+a}} = \tan \theta$.
$\frac{k(h+a) - k(a-h)}{(a-h)(h+a) + k^2} = \tan \theta$.
$\frac{2kh}{a^2 - h^2 + k^2} = \tan \theta$.
$2kh = (a^2 - h^2 + k^2) \tan \theta$.
$2kh \cot \theta = a^2 - h^2 + k^2$.
$h^2 - k^2 + 2hk \cot \theta = a^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 + 2xy \cot \theta = a^2$.

(A) Let the third vertex be $C(x_1, y_1)$.
The centroid $G$ of the triangle $ABC$ is given by:
$G = \left( \frac{x_1 + 2 - 2}{3}, \frac{y_1 - 3 + 1}{3} \right) = \left( \frac{x_1}{3}, \frac{y_1 - 2}{3} \right)$.
It is given that the centroid $G$ moves on the line $2x + 3y = 1$.
Substituting the coordinates of $G$ into the equation of the line:
$2\left( \frac{x_1}{3} \right) + 3\left( \frac{y_1 - 2}{3} \right) = 1$.
Multiplying the entire equation by $3$:
$2x_1 + 3(y_1 - 2) = 3$.
$2x_1 + 3y_1 - 6 = 3$.
$2x_1 + 3y_1 = 9$.
Thus,the locus of the vertex $C(x_1, y_1)$ is $2x + 3y = 9$.

(C) The given equation is $x(a + 2b) + y(a + 3b) = a + b$.
Rearranging the terms to group $a$ and $b$:
$x(a + 2b) + y(a + 3b) - (a + b) = 0$
$ax + 2bx + ay + 3by - a - b = 0$
$a(x + y - 1) + b(2x + 3y - 1) = 0$
For this equation to hold for all values of $a$ and $b$,the coefficients of $a$ and $b$ must be zero independently:
$x + y - 1 = 0 \implies x + y = 1$ $(i)$
$2x + 3y - 1 = 0 \implies 2x + 3y = 1$ $(ii)$
Multiplying equation $(i)$ by $2$,we get $2x + 2y = 2$ $(iii)$.
Subtracting $(iii)$ from $(ii)$:
$(2x + 3y) - (2x + 2y) = 1 - 2$
$y = -1$
Substituting $y = -1$ into equation $(i)$:
$x - 1 = 1 \implies x = 2$
Thus,the fixed point is $(2, -1)$.

(D) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the reciprocal of the sum of the intercepts is $1/p$,we have $\frac{1}{a+b} = \frac{1}{p}$,which implies $a+b = p$.
Substituting $b = p-a$ into the equation of the line:
$\frac{x}{a} + \frac{y}{p-a} = 1$
$x(p-a) + ay = a(p-a)$
$xp - xa + ay = ap - a^2$
$a^2 + a(y - x - p) + xp = 0$
For the line to pass through a fixed point regardless of the value of $a$,the equation must hold for all $a$. This implies the coefficients of the powers of $a$ must be zero.
However,the standard interpretation of this problem is that the sum of the reciprocals of the intercepts is constant,or the intercepts satisfy a specific relation. Re-evaluating the condition: if the sum of the intercepts is $p$,then $a+b=p$. The line $\frac{x}{a} + \frac{y}{p-a} = 1$ passes through $(p, p)$ because $\frac{p}{a} + \frac{p}{p-a} = \frac{p(p-a) + ap}{a(p-a)} = \frac{p^2 - ap + ap}{a(p-a)} = \frac{p^2}{a(p-a)}$. This does not equal $1$ for all $a$.
If the problem implies the sum of the reciprocals of the intercepts is $1/p$,i.e.,$\frac{1}{a} + \frac{1}{b} = \frac{1}{p}$,then $\frac{a+b}{ab} = \frac{1}{p}$,so $ab = p(a+b)$. The equation $\frac{x}{a} + \frac{y}{b} = 1$ becomes $bx + ay = ab = p(a+b) = pa + pb$. Rearranging gives $a(x-p) + b(y-p) = 0$. This line passes through the fixed point $(p, p)$.

(D) Let $m$ be the slope of the line $PQ$. The equation of the line is $y - 2 = m(x - 1)$.
The line meets the $X$-axis at $P$ where $y=0$,so $0 - 2 = m(x - 1) \implies x - 1 = -\frac{2}{m} \implies x = 1 - \frac{2}{m}$. Thus,$P = (1 - \frac{2}{m}, 0)$.
The line meets the $Y$-axis at $Q$ where $x=0$,so $y - 2 = m(0 - 1) \implies y = 2 - m$. Thus,$Q = (0, 2 - m)$.
The area $A$ of $\triangle OPQ = \frac{1}{2} |OP| |OQ| = \frac{1}{2} |(1 - \frac{2}{m})(2 - m)|$.
Since the line must form a triangle in the first quadrant for a minimum area,$1 - \frac{2}{m} < 0$ and $2 - m > 0$,implying $m < 0$.
$A = \frac{1}{2} |2 - m - \frac{4}{m} + 2| = \frac{1}{2} |4 - (m + \frac{4}{m})|$.
Since $m < 0$,let $m = -k$ where $k > 0$. Then $A = \frac{1}{2} |4 - (-k - \frac{4}{k})| = \frac{1}{2} (4 + k + \frac{4}{k})$.
To minimize $A$,we minimize $k + \frac{4}{k}$. By $AM$-$GM$ inequality,$k + \frac{4}{k} \geq 2\sqrt{k \cdot \frac{4}{k}} = 4$.
Equality holds when $k = \frac{4}{k} \implies k^2 = 4 \implies k = 2$.
Since $m = -k$,the slope $m = -2$.

(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1 \quad \dots(1)$ where $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \quad \dots(2)$.
Let $(h, k)$ be the foot of the perpendicular from the origin $(0, 0)$ to the line $(1)$.
The line passing through $(0, 0)$ and $(h, k)$ is perpendicular to the line $(1)$.
The slope of line $(1)$ is $m_1 = -\frac{b}{a}$.
The slope of the line passing through $(0, 0)$ and $(h, k)$ is $m_2 = \frac{k}{h}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1 \implies (-\frac{b}{a}) \times (\frac{k}{h}) = -1 \implies \frac{b}{a} = \frac{h}{k} \implies \frac{a}{h} = \frac{b}{k} = \lambda$.
Thus,$a = h\lambda$ and $b = k\lambda$.
Since $(h, k)$ lies on line $(1)$,$\frac{h}{h\lambda} + \frac{k}{k\lambda} = 1 \implies \frac{1}{\lambda} + \frac{1}{\lambda} = 1 \implies \lambda = 2$.
So,$a = 2h$ and $b = 2k$.
Substituting these into equation $(2)$,we get $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2} \implies \frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \implies \frac{h^2 + k^2}{4h^2k^2} = \frac{1}{c^2}$.
Wait,let's re-evaluate the perpendicular condition. The line is $\frac{x}{a} + \frac{y}{b} = 1$. The perpendicular line through origin is $bx - ay = 0$. The intersection $(h, k)$ satisfies $\frac{h}{a} + \frac{k}{b} = 1$ and $bh - ak = 0 \implies \frac{h}{a} = \frac{k}{b} = \frac{h^2+k^2}{ah+bk} = \frac{h^2+k^2}{a(h) + b(k)}$.
Actually,using the property of the foot of the perpendicular $(h, k)$ for $\frac{x}{a} + \frac{y}{b} = 1$: $h = \frac{a b^2}{a^2+b^2}$ and $k = \frac{a^2 b}{a^2+b^2}$.
Then $h^2 + k^2 = \frac{a^2 b^4 + a^4 b^2}{(a^2+b^2)^2} = \frac{a^2 b^2 (b^2 + a^2)}{(a^2+b^2)^2} = \frac{a^2 b^2}{a^2+b^2}$.
Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \implies \frac{a^2+b^2}{a^2b^2} = \frac{1}{c^2} \implies \frac{a^2b^2}{a^2+b^2} = c^2$.
Therefore,$h^2 + k^2 = c^2$. The locus is $x^2 + y^2 = c^2$.

(A) Given that $a, b, c$ are in Arithmetic Progression $(AP)$,we have $2b = a + c$,which implies $a - 2b + c = 0$.
Comparing this with the equation of the line $ax + by + c = 0$,we can rewrite the equation as $a(1) + b(x) + c = 0$ is not correct,let us substitute $c = 2b - a$ into the line equation:
$ax + by + (2b - a) = 0$
$a(x - 1) + b(y + 2) = 0$
For this equation to hold for all $a$ and $b$,the coefficients must be zero:
$x - 1 = 0 \implies x = 1$
$y + 2 = 0 \implies y = -2$
Thus,the fixed point is $(1, -2)$.

(A) Let $(x, y)$ be any point on the locus.
By the condition of equidistance,the distance from $(x, y)$ to $(a_1, b_1)$ equals the distance to $(a_2, b_2)$.
$(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2$
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 match the 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 the entire equation 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 $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,we get:
$c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$

(B) Let the equation of the variable line be $ax + by + c = 0$.
The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by $d = \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}$.
Given that the algebraic sum of the perpendicular distances from $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ is zero:
$\frac{2a + c}{\sqrt{a^2 + b^2}} + \frac{2b + c}{\sqrt{a^2 + b^2}} + \frac{a + b + c}{\sqrt{a^2 + b^2}} = 0$
Multiplying by $\sqrt{a^2 + b^2}$ (assuming $\sqrt{a^2 + b^2} \neq 0$):
$(2a + c) + (2b + c) + (a + b + c) = 0$
Combining like terms:
$3a + 3b + 3c = 0$
Dividing by $3$:
$a + b + c = 0$
This implies $c = -(a + b)$. Substituting this into the line equation $ax + by + c = 0$:
$ax + by - (a + b) = 0$
$a(x - 1) + b(y - 1) = 0$
This equation holds for all $a$ and $b$ if $x - 1 = 0$ and $y - 1 = 0$,which means $x = 1$ and $y = 1$.
Thus,all such lines pass through the fixed point $(1, 1)$.

(C) Given the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$ and $a + b = 10$.
The line intersects the $x$-axis at $A = (a, 0)$ and the $y$-axis at $B = (0, b)$.
Let the midpoint of $AB$ be $(x, y)$.
Then $x = \frac{a + 0}{2} = \frac{a}{2} \implies a = 2x$.
And $y = \frac{0 + b}{2} = \frac{b}{2} \implies b = 2y$.
Substitute $a$ and $b$ into the given condition $a + b = 10$:
$2x + 2y = 10$.
Dividing by $2$,we get $x + y = 5$.
Thus,the locus of the midpoint is $x + y = 5$.

(A) Given $A = (a, 0)$ and $B = (-a, 0)$.
Let $P = (x, y)$.
The condition is $PA = nPB$.
For $n = 1$,we have $PA = PB$.
Squaring both sides,$PA^2 = PB^2$.
$(x - a)^2 + (y - 0)^2 = (x - (-a))^2 + (y - 0)^2$.
$(x - a)^2 + y^2 = (x + a)^2 + y^2$.
$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 + y^2$.
$-2ax = 2ax$.
$4ax = 0$.
Since $a \neq 0$ (as $a \in (-\infty, 0)$),we get $x = 0$.
The equation $x = 0$ represents the $y$-axis,which is a straight line.

(B) Let the equation of the line passing through $(a, b)$ be $y - b = m(x - a)$.
For point $A$ (where $y = 0$): $-b = m(x - a) \implies x = a - \frac{b}{m}$. So,$A = (a - \frac{b}{m}, 0)$.
For point $B$ (where $x = 0$): $y - b = m(0 - a) \implies y = b - ma$. So,$B = (0, b - ma)$.
Let the intersection point of the lines parallel to the axes be $P(h, k)$.
Then $h = a - \frac{b}{m}$ and $k = b - ma$.
From the first equation,$\frac{b}{m} = a - h \implies m = \frac{b}{a - h}$.
Substitute $m$ into the second equation: $k = b - (\frac{b}{a - h})a$.
$k = \frac{b(a - h) - ab}{a - h} = \frac{ab - bh - ab}{a - h} = \frac{-bh}{a - h}$.
$k(a - h) = -bh \implies ak - kh = -bh \implies ak = kh - bh$.
Dividing by $hka$: $\frac{ak}{hka} = \frac{kh}{hka} - \frac{bh}{hka} \implies \frac{1}{h} = \frac{1}{a} - \frac{b}{ka}$.
Rearranging gives $\frac{a}{h} + \frac{b}{k} = 1$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{a}{x} + \frac{b}{y} = 1$.

(B) The lines are $L_1: x - 2y = 0$ and $L_2: 3x - y = 0$.
For the point $(a, a^2)$ to lie inside the angle,it must satisfy the inequalities formed by the lines such that it lies on the same side of $L_1$ as the origin (if applicable) or specifically between the rays.
Given $x > 0$,we have $a > 0$.
For the point to be below $y = 3x$,we need $a^2 < 3a$,which implies $a(a - 3) < 0$,so $0 < a < 3$.
For the point to be above $y = \frac{x}{2}$,we need $a^2 > \frac{a}{2}$,which implies $a(a - \frac{1}{2}) > 0$. Since $a > 0$,we have $a > \frac{1}{2}$.
Combining these,we get $\frac{1}{2} < a < 3$.

(B) Let the coordinates of $P$ be $(h, 0)$ and $Q$ be $(0, k)$.
Since the rectangle $OPRQ$ is completed,the coordinates of $R$ are $(h, k)$.
The equation of the line passing through $P(h, 0)$ and $Q(0, k)$ is given by the intercept form:
$\frac{x}{h} + \frac{y}{k} = 1$
Since this line passes through the fixed point $(2, 3)$,we have:
$\frac{2}{h} + \frac{3}{k} = 1$
To find the locus of $R(h, k)$,we replace $h$ with $x$ and $k$ with $y$:
$\frac{2}{x} + \frac{3}{y} = 1$
Multiplying both sides by $xy$,we get:
$2y + 3x = xy$
Thus,the locus of $R$ is $3x + 2y = xy$.