System of co-ordinates, Distance between two points, Section formulae Questions in English

(C) Let the point on the $x$-axis be $P(x, 0)$.
Given the distance between $P(x, 0)$ and $(2, 3)$ is $c$.
Using the distance formula: $\sqrt{(x-2)^2 + (0-3)^2} = c$.
Squaring both sides: $(x-2)^2 + 9 = c^2$.
Since $c < 3$,$c^2 < 9$.
Therefore,$(x-2)^2 = c^2 - 9$.
Since $c^2 < 9$,$c^2 - 9 < 0$.
Since the square of a real number cannot be negative,there is no real value of $x$ that satisfies this equation.
Thus,the number of such points is $0$.

(B) Let the point $P(8, 4)$ divide the line segment joining $A(5, -2)$ and $B(9, 6)$ in the ratio $k : 1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{k(9) + 1(5)}{k + 1}, \frac{k(6) + 1(-2)}{k + 1} \right)$
Equating the $x$-coordinate:
$8 = \frac{9k + 5}{k + 1}$
$8(k + 1) = 9k + 5$
$8k + 8 = 9k + 5$
$8 - 5 = 9k - 8k$
$k = 3$
Thus,the ratio $k : 1$ is $3 : 1$.

(D) The points $P, Q,$ and $R$ divide the line segment $AB$ into four equal parts. Thus,the ratio $AR:RB$ is $3:1$.
Using the section formula,if a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$,the coordinates are given by:
$x = \frac{mx_2 + nx_1}{m+n}, y = \frac{my_2 + ny_1}{m+n}$
For point $R$,$m=3$ and $n=1$,with $A(-6, 8)$ and $B(8, -6)$:
$x = \frac{3(8) + 1(-6)}{3+1} = \frac{24 - 6}{4} = \frac{18}{4} = \frac{9}{2}$
$y = \frac{3(-6) + 1(8)}{3+1} = \frac{-18 + 8}{4} = \frac{-10}{4} = -\frac{5}{2}$
Therefore,the coordinates of $R$ are $(\frac{9}{2}, -\frac{5}{2})$.

(C) The distance of a point $(x, y)$ from the $x$-axis is given by the absolute value of its $y$-coordinate,which is $|y|$.
Given the point $(6, 8)$,the $x$-coordinate is $6$ and the $y$-coordinate is $8$.
Therefore,the distance from the $x$-axis is $|8| = 8$ units.

(C) The formula for external division of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ is given by:
$P = (\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n})$
Given points are $(x_1, y_1) = (-3, -4)$ and $(x_2, y_2) = (-8, 7)$ with ratio $m : n = 7 : 5$.
Substituting the values:
$x = \frac{7(-8) - 5(-3)}{7 - 5} = \frac{-56 + 15}{2} = \frac{-41}{2}$
$y = \frac{7(7) - 5(-4)}{7 - 5} = \frac{49 + 20}{2} = \frac{69}{2}$
Thus,the coordinates are $(-41/2, 69/2)$.

(A) Let the point be $P(-3, 4)$.
The perpendicular distance from $P(-3, 4)$ to the $x$-axis is $|y| = |4| = 4$.
The perpendicular distance from $P(-3, 4)$ to the $y$-axis is $|x| = |-3| = 3$.
The sum of the squares of these lengths is $4^2 + 3^2 = 16 + 9 = 25$.
The square root of this sum is $\sqrt{25} = 5$.

(C) The polar coordinates are given as $(r, \theta) = (2, \pi /4)$.
To convert polar coordinates $(r, \theta)$ to Cartesian coordinates $(x, y)$,we use the formulas:
$x = r \cos \theta$
$y = r \sin \theta$
Substituting the given values:
$x = 2 \cos(\pi /4) = 2 \times (1 / \sqrt{2}) = \sqrt{2}$
$y = 2 \sin(\pi /4) = 2 \times (1 / \sqrt{2}) = \sqrt{2}$
Thus,the Cartesian coordinates are $(\sqrt{2}, \sqrt{2})$.

(C) Given $AC = 2BC$,which implies $\frac{AC}{BC} = \frac{2}{1}$.
Since $C$ lies on the line segment $AB$,it divides $AB$ internally in the ratio $m:n = 2:1$.
Using the internal section formula,the coordinates of $C(x, y)$ are given by:
$x = \frac{mx_2 + nx_1}{m+n} = \frac{2(2) + 1(-3)}{2+1} = \frac{4-3}{3} = \frac{1}{3}$
$y = \frac{my_2 + ny_1}{m+n} = \frac{2(1) + 1(4)}{2+1} = \frac{2+4}{3} = \frac{6}{3} = 2$
Thus,the coordinates of $C$ are $(\frac{1}{3}, 2)$.
Note: The provided options do not contain the correct result $(\frac{1}{3}, 2)$. However,if $C$ were to divide $AB$ externally in the ratio $2:1$,the coordinates would be:
$x = \frac{mx_2 - nx_1}{m-n} = \frac{2(2) - 1(-3)}{2-1} = \frac{4+3}{1} = 7$
$y = \frac{my_2 - ny_1}{m-n} = \frac{2(1) - 1(4)}{2-1} = \frac{2-4}{1} = -2$
This matches option $C$.

(C) In a right-angled triangle,the length of the median to the hypotenuse is equal to half the length of the hypotenuse.
First,calculate the length of the hypotenuse using the distance formula between $(0, 0)$ and $(3, 4)$:
$L = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The length of the median to the hypotenuse is half of the hypotenuse length:
$\text{Median} = \frac{1}{2} \times 5 = 2.5$.

(C) The coordinates of point $A$ using the section formula are $\left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} \right)$.
Given $B = (1, 5)$ and $C = (7, -2)$,the area of $\triangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 2$.
Substituting the coordinates: $\frac{1}{2} \left| \frac{3k - 5}{k + 1}(5 - (-2)) + 1(-2 - \frac{5k + 1}{k + 1}) + 7(\frac{5k + 1}{k + 1} - 5) \right| = 2$.
$\frac{1}{2} \left| \frac{7(3k - 5) - (2k + 2 + 5k + 1) + 7(5k + 1 - 5k - 5)}{k + 1} \right| = 2$.
$\frac{1}{2} \left| \frac{21k - 35 - 7k - 3 - 28}{k + 1} \right| = 2$.
$|14k - 66| = 4|k + 1|$.
Case $1$: $14k - 66 = 4k + 4$ $\Rightarrow 10k = 70$ $\Rightarrow k = 7$.
Case $2$: $14k - 66 = -4k - 4$ $\Rightarrow 18k = 62$ $\Rightarrow k = 31/9$.

(C) Given $AC = 2 BC$,we have $\frac{AC}{BC} = \frac{2}{1}$.
Since $C$ lies on $AB$ produced,$C$ divides the line segment $AB$ externally in the ratio $m:n = 2:1$.
The coordinates of a point dividing the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ externally in the ratio $m:n$ are given by $\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$.
Substituting $(x_1, y_1) = (-3, 4)$,$(x_2, y_2) = (2, 1)$,$m = 2$,and $n = 1$:
$x = \frac{2(2) - 1(-3)}{2-1} = \frac{4 + 3}{1} = 7$
$y = \frac{2(1) - 1(4)}{2-1} = \frac{2 - 4}{1} = -2$
Thus,the coordinates of $C$ are $(7, -2)$.

(A) The given points are $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$.
When $PQ$ is parallel to the $y$-axis,the $x$-coordinates are equal,so $x_{1} = x_{2}$.
The distance formula between two points is given by $d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$.
Substituting $x_{1} = x_{2}$ into the formula,we get $d = \sqrt{(x_{2}-x_{2})^{2} + (y_{2}-y_{1})^{2}}$.
$d = \sqrt{0^{2} + (y_{2}-y_{1})^{2}}$.
$d = \sqrt{(y_{2}-y_{1})^{2}} = |y_{2}-y_{1}|$.

(A) The given points are $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$.
Since $PQ$ is parallel to the $x$-axis,the $y$-coordinates of both points must be equal,i.e.,$y_{1} = y_{2}$.
The distance formula is $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$.
Substituting $y_{2} = y_{1}$ into the formula,we get $d = \sqrt{(x_{2}-x_{1})^2 + (y_{1}-y_{1})^2}$.
$d = \sqrt{(x_{2}-x_{1})^2 + 0^2}$.
$d = \sqrt{(x_{2}-x_{1})^2} = |x_{2}-x_{1}|$.

(A) Let $(a, 0)$ be the point on the $x$-axis that is equidistant from the points $P(7, 6)$ and $Q(3, 4)$.
According to the distance formula,the distance from $(a, 0)$ to $(7, 6)$ is equal to the distance from $(a, 0)$ to $(3, 4)$.
$\sqrt{(7-a)^2 + (6-0)^2} = \sqrt{(3-a)^2 + (4-0)^2}$
Squaring both sides:
$(7-a)^2 + 36 = (3-a)^2 + 16$
$(49 - 14a + a^2) + 36 = (9 - 6a + a^2) + 16$
$85 - 14a = 25 - 6a$
$85 - 25 = 14a - 6a$
$60 = 8a$
$a = \frac{60}{8} = \frac{15}{2}$
Thus,the required point on the $x$-axis is $(\frac{15}{2}, 0)$.

(B) The coordinates of the mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$ are calculated as:
$\left(\frac{0+8}{2}, \frac{-4+0}{2}\right) = (4, -2)$.
The slope $(m)$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$m = \frac{y_2 - y_1}{x_2 - x_1}$.
The line passes through the origin $(0, 0)$ and the mid-point $(4, -2)$.
Substituting these values into the slope formula:
$m = \frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$.
Thus,the slope of the line is $-\frac{1}{2}$.

(A) The $y$-coordinate of every point on the $x$-axis is $0$.
Therefore,the equation of the $x$-axis is $y=0$.
The $x$-coordinate of every point on the $y$-axis is $0$.
Therefore,the equation of the $y$-axis is $x=0$.

(C) The distance $d$ between two points $A(a_1, a_2)$ and $B(b_1, b_2)$ is given by $d = \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2}$.
Since $a_1, a_2, b_1, b_2$ are integers,let $x = |b_1 - a_1|$ and $y = |b_2 - a_2|$,where $x$ and $y$ are non-negative integers.
Thus,$d^2 = x^2 + y^2$.
We check the given options:
$A: \sqrt{65} = \sqrt{8^2 + 1^2}$,which is possible.
$B: \sqrt{74} = \sqrt{7^2 + 5^2}$,which is possible.
$C: \sqrt{83}$. We check if $83$ can be written as the sum of two squares. The squares less than $83$ are $0, 1, 4, 9, 16, 25, 36, 49, 64, 81$. No two of these sum to $83$.
$D: \sqrt{97} = \sqrt{9^2 + 4^2}$,which is possible.
Therefore,$\sqrt{83}$ is not a possible value.

(A) Given that the number of subsets of set $A$ is $2^m$ and the number of subsets of set $B$ is $2^n$.
According to the problem,$2^m - 2^n = 56$.
$2^n(2^{m-n} - 1) = 56 = 8 \times 7 = 2^3 \times 7$.
Comparing the powers of $2$,we get $2^n = 2^3$,which implies $n = 3$.
Also,$2^{m-n} - 1 = 7$,so $2^{m-n} = 8 = 2^3$.
Thus,$m - n = 3$,and substituting $n = 3$,we get $m = 6$.
The points are $P(6, 3)$ and $Q(-2, -3)$.
The distance $PQ = \sqrt{(6 - (-2))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.
Therefore,the correct option is $A$.

(C) Let $P = (x_1, y_1) = (-\sin(\beta - \alpha), -\cos \beta)$ and $Q = (x_2, y_2) = (\cos(\beta - \alpha), \sin \beta)$.
We observe that $R = (\cos(\beta - \alpha + \theta), \sin(\beta - \theta))$.
Using trigonometric identities,$\cos(\beta - \alpha + \theta) = \cos(\beta - \alpha)\cos \theta - \sin(\beta - \alpha)\sin \theta = x_2 \cos \theta + x_1 \sin \theta$.
Similarly,$\sin(\beta - \theta) = \sin \beta \cos \theta - \cos \beta \sin \theta = y_2 \cos \theta + y_1 \sin \theta$.
Thus,$R = (x_2 \cos \theta + x_1 \sin \theta, y_2 \cos \theta + y_1 \sin \theta)$.
This shows that $R$ divides the line segment $PQ$ internally in the ratio $\sin \theta : \cos \theta$.
Since $0 < \theta < \frac{\pi}{4}$,both $\sin \theta$ and $\cos \theta$ are positive,so $R$ lies on the line segment $PQ$ (or $QP$).

(B) Let the points be $A(1, 1)$ and $B(2, 4)$. The point $P(x, y)$ divides $AB$ in the ratio $m:n = 3:2$.
Using the section formula,$x = \frac{mx_2 + nx_1}{m+n} = \frac{3(2) + 2(1)}{3+2} = \frac{6+2}{5} = \frac{8}{5}$.
$y = \frac{my_2 + ny_1}{m+n} = \frac{3(4) + 2(1)}{3+2} = \frac{12+2}{5} = \frac{14}{5}$.
The point $P$ is $(\frac{8}{5}, \frac{14}{5})$.
Since the line $2x + y = k$ passes through $P$,we substitute the coordinates of $P$ into the equation:
$2(\frac{8}{5}) + \frac{14}{5} = k \implies \frac{16}{5} + \frac{14}{5} = k \implies k = \frac{30}{5} = 6$.
Now,we need to find the ratio $(k+1):(k-1)$.
Substituting $k=6$,we get $(6+1):(6-1) = 7:5$.
Thus,the value is $\frac{7}{5}$.

(D) The relation between polar coordinates $(r, \theta)$ and Cartesian coordinates $(x, y)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$.
Given $r = \sqrt{2}$ and $\theta = \frac{\pi}{4}$.
$x = \sqrt{2} \cos \left(\frac{\pi}{4}\right) = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$.
$y = \sqrt{2} \sin \left(\frac{\pi}{4}\right) = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$.
Therefore,the Cartesian coordinates are $(1, 1)$.

(A) The relationship between polar coordinates $(r, \theta)$ and Cartesian coordinates $(x, y)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$.
Given $r = 2$ and $\theta = \frac{\pi}{4}$.
$x = 2 \cos \left(\frac{\pi}{4}\right) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
$y = 2 \sin \left(\frac{\pi}{4}\right) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
Thus,the Cartesian coordinates are $(\sqrt{2}, \sqrt{2})$.

(A) We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$.
Since the point $(-2, -2)$ lies in the $III$ quadrant,the angle $\theta$ is given by $\tan \theta = \frac{y}{x} = \frac{-2}{-2} = 1$.
Since the point is in the $III$ quadrant,$\theta = \pi + \tan^{-1}(1) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$.
Therefore,the polar coordinates $(r, \theta)$ are $(2 \sqrt{2}, \frac{5 \pi}{4})$.

(A) The points $P, Q, R$ divide the line segment $AB$ into $4$ equal parts. Thus,$AP = PQ = QR = RB = k$ (let).
$P$ divides $AB$ in the ratio $1:3$,$Q$ divides $AB$ in the ratio $2:2$ (or $1:1$),and $R$ divides $AB$ in the ratio $3:1$.
Since $Q$ is the midpoint of $AB$,its coordinates are $\left( \frac{2+6}{2}, \frac{-7+5}{2} \right) = (4, -1)$.
$P$ is the midpoint of $AQ$. The coordinates of $A$ are $(2, -7)$ and $Q$ are $(4, -1)$.
Coordinates of $P = \left( \frac{2+4}{2}, \frac{-7-1}{2} \right) = (3, -4)$.
$R$ is the midpoint of $QB$. The coordinates of $Q$ are $(4, -1)$ and $B$ are $(6, 5)$.
Coordinates of $R = \left( \frac{4+6}{2}, \frac{-1+5}{2} \right) = (5, 2)$.
The midpoint of $PR$ is $\left( \frac{3+5}{2}, \frac{-4+2}{2} \right) = (4, -1)$.

(C) Let the line $3x + 4y = 6$ divide the line segment joining the points $P(2, -1)$ and $Q(1, 1)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of intersection are $\left(\frac{1(2) + k(1)}{k+1}, \frac{1(-1) + k(1)}{k+1}\right) = \left(\frac{2+k}{k+1}, \frac{k-1}{k+1}\right)$.
Since this point lies on the line $3x + 4y = 6$,we substitute these coordinates into the equation:
$3\left(\frac{2+k}{k+1}\right) + 4\left(\frac{k-1}{k+1}\right) = 6$.
Multiplying by $(k+1)$,we get:
$3(2+k) + 4(k-1) = 6(k+1)$.
$6 + 3k + 4k - 4 = 6k + 6$.
$7k + 2 = 6k + 6$.
$k = 4$.
Thus,the ratio is $4:1$.

(A) The point $C$ divides the segment $AB$ in the ratio $a:b$. Using the section formula,the coordinates of $C$ are $\left(\frac{-2a + b}{a + b}, \frac{a + 2b}{a + b}\right)$.
Since $C$ lies on the line $2x + y - 3 = 0$,we have:
$2\left(\frac{-2a + b}{a + b}\right) + \left(\frac{a + 2b}{a + b}\right) - 3 = 0$
$-4a + 2b + a + 2b - 3(a + b) = 0$
$-3a + 4b - 3a - 3b = 0$
$b = 6a \Rightarrow \frac{b}{a} = 6$.
Now,substitute $\frac{b}{a} = 6$ into the coordinates of $P$ and $Q$:
$P = \left(\frac{6}{3}, -3\right) = (2, -3)$
$Q = \left(-3, -\frac{6}{3}\right) = (-3, -2)$
Point $C$ is $\left(\frac{-2a + 6a}{a + 6a}, \frac{a + 12a}{a + 6a}\right) = \left(\frac{4a}{7a}, \frac{13a}{7a}\right) = \left(\frac{4}{7}, \frac{13}{7}\right)$.
Let $C$ divide $PQ$ in the ratio $p:q$. Using the section formula for $x$-coordinate:
$\frac{-3p + 2q}{p + q} = \frac{4}{7}$
$-21p + 14q = 4p + 4q$
$10q = 25p \Rightarrow \frac{p}{q} = \frac{10}{25} = \frac{2}{5}$.
Then,$\frac{p}{q} + \frac{q}{p} = \frac{2}{5} + \frac{5}{2} = \frac{4 + 25}{10} = \frac{29}{10}$.

(A) Let the point $P$ divide the segment $AB$ in the ratio $m:n$. By the section formula,the coordinates of $P$ are $\left(\frac{3m - 2n}{m + n}, \frac{-2m + 3n}{m + n}\right)$.
Since $P$ lies on the line $2x - 3y + 4 = 0$,we have $2\left(\frac{3m - 2n}{m + n}\right) - 3\left(\frac{-2m + 3n}{m + n}\right) + 4 = 0$.
Multiplying by $(m + n)$,we get $6m - 4n + 6m - 9n + 4m + 4n = 0$,which simplifies to $16m - 9n = 0$,so $\frac{m}{n} = \frac{9}{16}$.
We need to find the point dividing $AB$ in the ratio $k = \frac{-4m}{3n} = \frac{-4}{3} \times \frac{9}{16} = -\frac{3}{4}$.
Using the section formula for ratio $k = -\frac{3}{4}$ with $A(-2, 3)$ and $B(3, -2)$:
$x = \frac{k x_2 + x_1}{k + 1} = \frac{-\frac{3}{4}(3) + (-2)}{-\frac{3}{4} + 1} = \frac{-\frac{9}{4} - 2}{\frac{1}{4}} = -9 - 8 = -17$.
$y = \frac{k y_2 + y_1}{k + 1} = \frac{-\frac{3}{4}(-2) + 3}{-\frac{3}{4} + 1} = \frac{\frac{3}{2} + 3}{\frac{1}{4}} = 4 \times \frac{9}{2} = 18$.
Thus,the point is $(-17, 18)$.

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The coordinates of the points are $A(a, 0)$ and $B(0, b)$.
Given that the point $P(2, 3)$ divides the line segment $AB$ in the ratio $2:3$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left(\frac{2 \cdot 0 + 3 \cdot a}{2 + 3}, \frac{2 \cdot b + 3 \cdot 0}{2 + 3}\right) = \left(\frac{3a}{5}, \frac{2b}{5}\right)$.
Equating the coordinates with $(2, 3)$:
$\frac{3a}{5} = 2$ $\Rightarrow 3a = 10$ $\Rightarrow a = \frac{10}{3}$.
$\frac{2b}{5} = 3$ $\Rightarrow 2b = 15$ $\Rightarrow b = \frac{15}{2}$.
The product of the intercepts is $a \cdot b = \left(\frac{10}{3}\right) \cdot \left(\frac{15}{2}\right) = 5 \cdot 5 = 25$.

(C) The point $N(x, y)$ divides the line segment $AB$ externally in the ratio $b: a$.
Using the external section formula,the coordinates of $N$ are given by:
$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$
$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$
From these,we have:
$\frac{x}{ab} = \frac{\cos \beta - \cos \alpha}{b - a} \implies \frac{b - a}{ab} = \frac{\cos \beta - \cos \alpha}{x}$
$\frac{y}{ab} = \frac{\sin \beta - \sin \alpha}{b - a} \implies \frac{b - a}{ab} = \frac{\sin \beta - \sin \alpha}{y}$
Equating the two expressions for $\frac{b - a}{ab}$:
$\frac{\cos \beta - \cos \alpha}{x} = \frac{\sin \beta - \sin \alpha}{y}$
$y(\cos \beta - \cos \alpha) = x(\sin \beta - \sin \alpha)$
Using the sum-to-product formulas:
$y \left( -2 \sin \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} \right) = x \left( 2 \cos \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} \right)$
Assuming $\sin \frac{\beta - \alpha}{2} \neq 0$,we divide both sides by $2 \sin \frac{\beta - \alpha}{2}$:
$-y \sin \frac{\alpha + \beta}{2} = x \cos \frac{\alpha + \beta}{2}$
$x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = 0$

(A) Let the points be $A = (a \cos \theta, a \sin \theta)$ and $B = (a \cos \phi, a \sin \phi)$.
Given the distance $AB = 2a$.
Using the distance formula:
$AB^2 = (a \cos \theta - a \cos \phi)^2 + (a \sin \theta - a \sin \phi)^2 = (2a)^2$
$a^2(\cos^2 \theta + \cos^2 \phi - 2 \cos \theta \cos \phi + \sin^2 \theta + \sin^2 \phi - 2 \sin \theta \sin \phi) = 4a^2$
$a^2(2 - 2(\cos \theta \cos \phi + \sin \theta \sin \phi)) = 4a^2$
$2 - 2 \cos(\theta - \phi) = 4$
$-2 \cos(\theta - \phi) = 2$
$\cos(\theta - \phi) = -1$
Since $\cos(\theta - \phi) = -1$,we have $\theta - \phi = (2n + 1)\pi = 2n\pi + \pi$ for $n \in Z$.
Therefore,$\theta = 2n\pi + \pi + \phi$.

(D) Let the ratio be $m:n = k:1$. The point $(a, b)$ divides the segment joining $(2, -1)$ and $(1, -4)$ in the ratio $k:1$.
Using the section formula,$a = \frac{k(1) + 1(2)}{k+1} = \frac{k+2}{k+1}$ and $b = \frac{k(-4) + 1(-1)}{k+1} = \frac{-4k-1}{k+1}$.
Since $(a, b)$ lies on the line $2x - y - 4 = 0$,we have $2(\frac{k+2}{k+1}) - (\frac{-4k-1}{k+1}) - 4 = 0$.
Multiplying by $(k+1)$,we get $2k + 4 + 4k + 1 - 4(k+1) = 0$,which simplifies to $6k + 5 - 4k - 4 = 0$,so $2k + 1 = 0$,giving $k = -\frac{1}{2}$.
Thus,$\frac{m}{n} = -\frac{1}{2}$.
Substituting $k = -\frac{1}{2}$ into the coordinates: $a = \frac{-0.5+2}{-0.5+1} = \frac{1.5}{0.5} = 3$ and $b = \frac{-4(-0.5)-1}{-0.5+1} = \frac{2-1}{0.5} = 2$.
Finally,$4(a - b(\frac{m}{n})^2) = 4(3 - 2(-\frac{1}{2})^2) = 4(3 - 2(\frac{1}{4})) = 4(3 - 0.5) = 4(2.5) = 10$.

(D) Let the points be $A(3, 5)$ and $B(4, 6)$. The line $L \equiv 6x + 3y + k = 0$ divides $AB$ in the ratio $m:n = -5: 4$.
Using the section formula,the point of intersection $Q$ is given by $\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) = \left( \frac{-5(4) + 4(3)}{-5 + 4}, \frac{-5(6) + 4(5)}{-5 + 4} \right) = \left( \frac{-20 + 12}{-1}, \frac{-30 + 20}{-1} \right) = (8, 10)$.
Since $Q(8, 10)$ lies on $L = 0$,we have $6(8) + 3(10) + k = 0 \implies 48 + 30 + k = 0 \implies k = -78$.
So,the line $L$ is $6x + 3y - 78 = 0$,which simplifies to $2x + y - 26 = 0$.
The point $P(g, h)$ is the intersection of $2x + y = 26$ and $x - y = -1$.
Adding the two equations: $(2x + y) + (x - y) = 26 - 1 \implies 3x = 25 \implies g = \frac{25}{3}$.
Substituting $g$ into $x - y = -1$: $h = g + 1 = \frac{25}{3} + 1 = \frac{28}{3}$.
Checking the options: $g + 1 = \frac{25}{3} + 1 = \frac{28}{3} = h$. Thus,$h = g + 1$.

(A) The distance $PQ$ between the two particles is given by the distance formula:
$PQ = \sqrt{((t+1) - t)^2 + ((t^3 - 6t - 6) - (t^3 - 16t - 3))^2}$
$PQ = \sqrt{(1)^2 + (t^3 - 6t - 6 - t^3 + 16t + 3)^2}$
$PQ = \sqrt{1 + (10t - 3)^2}$
To find the minimum distance,we minimize $PQ^2 = 1 + (10t - 3)^2$.
Since $(10t - 3)^2 \geq 0$ for all real $t$,the minimum value of $(10t - 3)^2$ is $0$ when $10t - 3 = 0$,i.e.,$t = 0.3$.
Thus,the minimum distance is $\sqrt{1 + 0} = 1$.

(A) The polar coordinate of point $P$ is given as $\left(2, -\frac{\pi}{4}\right)$.
Since the line segment $PQ$ is bisected perpendicularly by the initial line (the $X$-axis),point $Q$ must be the reflection of point $P$ across the $X$-axis.
In polar coordinates $(r, \theta)$,reflecting a point across the initial line changes the sign of the angle $\theta$ while keeping $r$ constant.
Therefore,the reflection of $\left(2, -\frac{\pi}{4}\right)$ across the $X$-axis is $\left(2, -\left(-\frac{\pi}{4}\right)\right) = \left(2, \frac{\pi}{4}\right)$.
Thus,the coordinates of $Q$ are $\left(2, \frac{\pi}{4}\right)$.

(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.
Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$,the coordinates of $M$ are given by the section formula for external division:
$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$
$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$
Now,consider the expression $E = x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$.
Substituting $x$ and $y$:
$E = \frac{ab}{b-a} [(\cos \beta - \cos \alpha) \cos \frac{\alpha+\beta}{2} + (\sin \beta - \sin \alpha) \sin \frac{\alpha+\beta}{2}]$
Using the identity $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$E = \frac{ab}{b-a} [(-2 \sin \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \cos \frac{\alpha+\beta}{2} + (2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \sin \frac{\alpha+\beta}{2}]$
$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [-\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} + \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha+\beta}{2}]$
$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [0] = 0$.

(B) Let the line $3x + y - 9 = 0$ divide the line segment joining $A(1, 3)$ and $B(2, 7)$ in the ratio $k: 1$.
Using the section formula,the coordinates of the dividing point are $(\frac{2k+1}{k+1}, \frac{7k+3}{k+1})$.
Since this point lies on the line $3x + y = 9$,we have:
$3(\frac{2k+1}{k+1}) + (\frac{7k+3}{k+1}) = 9$
$6k + 3 + 7k + 3 = 9(k + 1)$
$13k + 6 = 9k + 9$
$4k = 3$
$k = \frac{3}{4}$.
Since $k > 0$,the division is internal in the ratio $3: 4$.

(C) Given $A \equiv (2, 4)$.
Since $C$ is the reflection of $A$ in the $x$-axis,the coordinates of $C$ are $(2, -4)$.
Since $B$ is the reflection of $C$ in the $y$-axis,the coordinates of $B$ are $(-2, -4)$.
Now,we calculate the distance $|AB|$ using the distance formula:
$|AB| = \sqrt{(2 - (-2))^2 + (4 - (-4))^2}$
$|AB| = \sqrt{(2 + 2)^2 + (4 + 4)^2}$
$|AB| = \sqrt{4^2 + 8^2}$
$|AB| = \sqrt{16 + 64}$
$|AB| = \sqrt{80}$
$|AB| = \sqrt{16 \times 5} = 4 \sqrt{5}$

(A) Given that $C$ divides the line segment joining $A(-3, 4)$ and $B(2, 1)$ in the ratio $AC : BC = 2 : 1$.
Using the section formula,the coordinates of $C(x, y)$ are given by:
$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2(2) + 1(-3)}{2 + 1} = \frac{4 - 3}{3} = \frac{1}{3}$
$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{2(1) + 1(4)}{2 + 1} = \frac{2 + 4}{3} = \frac{6}{3} = 2$
Thus,the coordinates of $C$ are $\left(\frac{1}{3}, 2\right)$.