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

(B) Let $O$ be the origin $(0, 0)$. The points are $A(a, b)$ and $B(c, d)$.
Using the distance formula,we have:
$(OA)^2 = a^2 + b^2$
$(OB)^2 = c^2 + d^2$
$(AB)^2 = (a - c)^2 + (b - d)^2$
In $\triangle AOB$,by the Law of Cosines:
$\cos \theta = \frac{(OA)^2 + (OB)^2 - (AB)^2}{2(OA)(OB)}$
Substituting the values:
$\cos \theta = \frac{(a^2 + b^2) + (c^2 + d^2) - [(a - c)^2 + (b - d)^2]}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{a^2 + b^2 + c^2 + d^2 - (a^2 - 2ac + c^2 + b^2 - 2bd + d^2)}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{2ac + 2bd}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
Thus,the correct option is $B$.

(C) Let the coordinates of the first point be $P(0, 1)$ and the second point be $Q(x, -3)$.
Given that the distance between $P$ and $Q$ is $5$.
Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$5 = \sqrt{(x - 0)^2 + (-3 - 1)^2}$.
$5 = \sqrt{x^2 + (-4)^2}$.
$5 = \sqrt{x^2 + 16}$.
Squaring both sides: $25 = x^2 + 16$.
$x^2 = 25 - 16 = 9$.
$x = \pm 3$.
Therefore,the abscissa of the other point is $\pm 3$.

(B) In a Cartesian coordinate system,any point $P$ in the plane is represented as $(x, y)$.
For any point lying on the $x$-axis,the vertical distance from the $x$-axis is zero.
Therefore,the $y$-coordinate of every point on the $x$-axis is $0$.
Thus,the common property is $y = 0$.

(D) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = d$.
Given points are $(a, 2)$ and $(3, 4)$ with distance $d = 8$.
So,$\sqrt{(3 - a)^2 + (4 - 2)^2} = 8$.
Squaring both sides,we get $(3 - a)^2 + (2)^2 = 8^2$.
$(3 - a)^2 + 4 = 64$.
$(3 - a)^2 = 60$.
$3 - a = \pm \sqrt{60} = \pm 2\sqrt{15}$.
$a = 3 \pm 2\sqrt{15}$.

(A) Given $AB = BC$,we use the distance formula: $\sqrt{(6-1)^2 + (-1-3)^2} = \sqrt{(x-1)^2 + (8-3)^2}$.
Squaring both sides: $(5)^2 + (-4)^2 = (x-1)^2 + (5)^2$.
$25 + 16 = (x-1)^2 + 25$.
$16 = (x-1)^2$.
Taking the square root: $x-1 = \pm 4$.
Case $1$: $x-1 = 4 \Rightarrow x = 5$.
Case $2$: $x-1 = -4 \Rightarrow x = -3$.
Therefore,$x = 5, -3$.

(D) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the given points:
$d = \sqrt{(a \cos \beta - a \cos \alpha)^2 + (a \sin \beta - a \sin \alpha)^2}$
$d = \sqrt{a^2(\cos^2 \beta + \cos^2 \alpha - 2 \cos \alpha \cos \beta) + a^2(\sin^2 \beta + \sin^2 \alpha - 2 \sin \alpha \sin \beta)}$
$d = a \sqrt{(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)}$
Using trigonometric identities $\cos^2 \theta + \sin^2 \theta = 1$ and $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:
$d = a \sqrt{1 + 1 - 2 \cos(\alpha - \beta)}$
$d = a \sqrt{2(1 - \cos(\alpha - \beta))}$
Using $1 - \cos \theta = 2 \sin^2(\theta/2)$:
$d = a \sqrt{2 \times 2 \sin^2 \left(\frac{\alpha - \beta}{2}\right)}$
$d = 2a \left| \sin \left(\frac{\alpha - \beta}{2}\right) \right|$

(B) Let the point on the $y$-axis be $P(0, y)$.
Since $P$ is equidistant from $A(3, 2)$ and $B(-1, 3)$,we have $PA^2 = PB^2$.
$(0 - 3)^2 + (y - 2)^2 = (0 - (-1))^2 + (y - 3)^2$
$9 + y^2 - 4y + 4 = 1 + y^2 - 6y + 9$
$13 - 4y = 10 - 6y$
$2y = -3$
$y = -\frac{3}{2}$
Therefore,the required point is $(0, -\frac{3}{2})$.

(C) Let the vertices be $O(0, 0)$,$A(\cos \alpha, \sin \alpha)$,and $B(-\sin \alpha, \cos \alpha)$.
The square of the distance $OA$ is given by $OA^2 = (\cos \alpha - 0)^2 + (\sin \alpha - 0)^2 = \cos^2 \alpha + \sin^2 \alpha = 1$.
The square of the distance $OB$ is given by $OB^2 = (-\sin \alpha - 0)^2 + (\cos \alpha - 0)^2 = \sin^2 \alpha + \cos^2 \alpha = 1$.
Therefore,$OA^2 + OB^2 = 1 + 1 = 2$.

(B) The distance $d$ of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $d = \sqrt{x^2 + y^2}$.
Substituting the given coordinates $(b \cos \theta, b \sin \theta)$:
$d = \sqrt{(b \cos \theta)^2 + (b \sin \theta)^2}$
$d = \sqrt{b^2 \cos^2 \theta + b^2 \sin^2 \theta}$
$d = \sqrt{b^2(\cos^2 \theta + \sin^2 \theta)}$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$d = \sqrt{b^2(1)} = b$.

(A) Let the points be $P(a \sin \theta, 0)$ and $Q(0, a \cos \theta)$.
The midpoint $M$ of the line segment $PQ$ is given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Substituting the coordinates,$M = \left( \frac{a \sin \theta + 0}{2}, \frac{0 + a \cos \theta}{2} \right) = \left( \frac{a \sin \theta}{2}, \frac{a \cos \theta}{2} \right)$.
The distance of $M$ from the origin $(0, 0)$ is $\sqrt{\left( \frac{a \sin \theta}{2} - 0 \right)^2 + \left( \frac{a \cos \theta}{2} - 0 \right)^2}$.
$= \sqrt{\frac{a^2 \sin^2 \theta}{4} + \frac{a^2 \cos^2 \theta}{4}} = \sqrt{\frac{a^2}{4} (\sin^2 \theta + \cos^2 \theta)}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,the distance is $\sqrt{\frac{a^2}{4}} = \frac{a}{2}$.

(D) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(7, 5)$ and $(3, 2)$.
Substituting the values: $d = \sqrt{(3 - 7)^2 + (2 - 5)^2}$.
$d = \sqrt{(-4)^2 + (-3)^2}$.
$d = \sqrt{16 + 9}$.
$d = \sqrt{25} = 5 \text{ units}$.
Therefore,the correct option is $D$.

(A) The midpoint formula for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Given the points $(5, a)$ and $(b, 7)$ and the midpoint $(3, 5)$,we have:
$\frac{5 + b}{2} = 3 \implies 5 + b = 6 \implies b = 1$.
$\frac{a + 7}{2} = 5 \implies a + 7 = 10 \implies a = 3$.
Therefore,$(a, b) = (3, 1)$.

(B) Let the ratio in which the $x$-axis divides the line segment joining the points $A(2, -3)$ and $B(5, 6)$ be $k : 1$.
Since the point lies on the $x$-axis,its $y$-coordinate must be $0$.
Using the section formula,the $y$-coordinate of the dividing point is given by $\frac{k(6) + 1(-3)}{k + 1} = 0$.
Solving for $k$: $6k - 3 = 0$ $\Rightarrow 6k = 3$ $\Rightarrow k = \frac{3}{6} = \frac{1}{2}$.
Thus,the ratio is $\frac{1}{2} : 1$,which is $1 : 2$.
Alternatively,for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ divided by the $x$-axis,the ratio is $-\frac{y_1}{y_2} = -\frac{-3}{6} = \frac{3}{6} = 1 : 2$.

(A) The section formula for external division of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ is given by $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right)$.
Given points are $(x_1, y_1) = (a + b, a - b)$ and $(x_2, y_2) = (a - b, a + b)$ with ratio $m:n = a:b$.
Applying the formula for the $x$-coordinate:
$x = \frac{a(a - b) - b(a + b)}{a - b} = \frac{a^2 - ab - ab - b^2}{a - b} = \frac{a^2 - 2ab - b^2}{a - b}$.
Applying the formula for the $y$-coordinate:
$y = \frac{a(a + b) - b(a - b)}{a - b} = \frac{a^2 + ab - ab + b^2}{a - b} = \frac{a^2 + b^2}{a - b}$.
Thus,the point is $\left( \frac{a^2 - 2ab - b^2}{a - b}, \frac{a^2 + b^2}{a - b} \right)$.

(A) The coordinates of $D$ which divides $AB$ in the ratio $l : k$ are given by the section formula:
$D = \left( \frac{l x_2 + k x_1}{l + k}, \frac{l y_2 + k y_1}{l + k} \right)$
Now,$P$ divides the line segment $DC$ in the ratio $m : (k + l)$.
Using the section formula for $P$ with $D = (x_D, y_D)$ and $C = (x_3, y_3)$:
$P = \left( \frac{m x_3 + (k + l) x_D}{m + k + l}, \frac{m y_3 + (k + l) y_D}{m + k + l} \right)$
Substituting the coordinates of $D$:
$x_P = \frac{m x_3 + (k + l) \left( \frac{l x_2 + k x_1}{l + k} \right)}{k + l + m} = \frac{m x_3 + l x_2 + k x_1}{k + l + m}$
Similarly,$y_P = \frac{m y_3 + l y_2 + k y_1}{k + l + m}$
Thus,the coordinates of $P$ are $\left( \frac{k x_1 + l x_2 + m x_3}{k + l + m}, \frac{k y_1 + l y_2 + m y_3}{k + l + m} \right)$.

(A) Let the points be $A(0, 0)$ and $D(9, 12)$. Let $B$ and $C$ be the points that trisect the line segment $AD$.
$(i)$ Point $B$ divides $AD$ in the ratio $1 : 2$.
Using the section formula,the coordinates of $B$ are:
$x = \frac{1(9) + 2(0)}{1 + 2} = \frac{9}{3} = 3$
$y = \frac{1(12) + 2(0)}{1 + 2} = \frac{12}{3} = 4$
So,$B = (3, 4)$.
$(ii)$ Point $C$ divides $AD$ in the ratio $2 : 1$.
Using the section formula,the coordinates of $C$ are:
$x = \frac{2(9) + 1(0)}{2 + 1} = \frac{18}{3} = 6$
$y = \frac{2(12) + 1(0)}{2 + 1} = \frac{24}{3} = 8$
So,$C = (6, 8)$.
Thus,the points are $(3, 4)$ and $(6, 8)$.

(B) Let the line $x + y - 4 = 0$ divide the line segment joining $A(-1, 1)$ and $B(5, 7)$ in the ratio $k : 1$.
Using the section formula,the coordinates of the point of division are $\left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right)$.
Since this point lies on the line $x + y = 4$,we have:
$\frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$
$5k - 1 + 7k + 1 = 4(k + 1)$
$12k = 4k + 4$
$8k = 4$
$k = \frac{4}{8} = \frac{1}{2}$.
Thus,the ratio is $1 : 2$.

(A) Let the points be $A(-2, 3)$ and $B(3, -1)$.
Let $C$ be the point of trisection nearer to $A$. Then $C$ divides the line segment $AB$ in the ratio $1:2$.
Using the section formula,the coordinates of $C$ are given by:
$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{1(3) + 2(-2)}{1 + 2} = \frac{3 - 4}{3} = -\frac{1}{3}$
$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{1(-1) + 2(3)}{1 + 2} = \frac{-1 + 6}{3} = \frac{5}{3}$
Thus,the coordinates of $C$ are $\left( -\frac{1}{3}, \frac{5}{3} \right)$.

(B) Let the vertices of the parallelogram be $A(-1, -6)$,$B(2, -5)$,$C(7, 2)$,and $D(x, y)$.
Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
The midpoint of $AC$ is $(\frac{-1+7}{2}, \frac{-6+2}{2}) = (3, -2)$.
The midpoint of $BD$ is $(\frac{2+x}{2}, \frac{-5+y}{2})$.
Equating the midpoints:
$\frac{2+x}{2} = 3$ $\Rightarrow 2+x = 6$ $\Rightarrow x = 4$
$\frac{-5+y}{2} = -2$ $\Rightarrow -5+y = -4$ $\Rightarrow y = 1$
Therefore,the fourth vertex is $(4, 1)$.

(A) Given that $P$ and $Q$ divide the line segment $AB$ into three equal parts such that $AP = PQ = QB$.
This implies that $P$ and $Q$ are the points of trisection of the line segment $AB$.
The mid-point of $PQ$ is the same as the mid-point of $AB$ because $P$ and $Q$ are symmetric with respect to the center of the segment $AB$.
The coordinates of the mid-point of $AB$ are given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Substituting the coordinates $A(-2, 5)$ and $B(3, 1)$:
Mid-point $= \left( \frac{-2 + 3}{2}, \frac{5 + 1}{2} \right) = \left( \frac{1}{2}, \frac{6}{2} \right) = \left( \frac{1}{2}, 3 \right)$.

(C) Let $A(3, -2)$ and $B(-3, -4)$ be the endpoints of the line segment,and let $C$ and $D$ be the points of trisection.
$C$ divides $AB$ in the ratio $1:2$. Using the section formula,the coordinates of $C$ are:
$\left( \frac{1(-3) + 2(3)}{1+2}, \frac{1(-4) + 2(-2)}{1+2} \right) = \left( \frac{3}{3}, \frac{-8}{3} \right) = \left( 1, - \frac{8}{3} \right)$.
$D$ divides $AB$ in the ratio $2:1$. Using the section formula,the coordinates of $D$ are:
$\left( \frac{2(-3) + 1(3)}{2+1}, \frac{2(-4) + 1(-2)}{2+1} \right) = \left( \frac{-3}{3}, \frac{-10}{3} \right) = \left( -1, - \frac{10}{3} \right)$.

(C) Using the section formula for internal division,the coordinates of the point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ are given by $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$.
Here,$(x_1, y_1) = (4, -2)$,$(x_2, y_2) = (8, 6)$,$m = 7$,and $n = 5$.
$x = \frac{7(8) + 5(4)}{7 + 5} = \frac{56 + 20}{12} = \frac{76}{12} = \frac{19}{3}$.
$y = \frac{7(6) + 5(-2)}{7 + 5} = \frac{42 - 10}{12} = \frac{32}{12} = \frac{8}{3}$.
Thus,the coordinates are $\left( \frac{19}{3}, \frac{8}{3} \right)$.

(C) Let the $y$-axis divide the line segment joining $A(-3, -4)$ and $B(1, -2)$ in the ratio $k : 1$ at point $P(0, y)$.
Using the section formula,the $x$-coordinate of point $P$ is given by:
$x = \frac{m x_2 + n x_1}{m + n}$
Since the point lies on the $y$-axis,its $x$-coordinate is $0$:
$0 = \frac{k(1) + 1(-3)}{k + 1}$
$0 = k - 3$
$k = 3$
Therefore,the required ratio is $3 : 1$.

(A) Let the ratio in which the $y$-axis divides the line segment joining the points $(x_1, y_1) = (2, -3)$ and $(x_2, y_2) = (-5, 6)$ be $k : 1$.
The coordinates of the point of division on the $y$-axis are given by $\left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right)$.
Since the point lies on the $y$-axis,its $x$-coordinate must be $0$.
Therefore,$\frac{k(-5) + 2}{k+1} = 0$.
$-5k + 2 = 0 \implies 5k = 2 \implies k = \frac{2}{5}$.
Thus,the required ratio is $2 : 5$.

(D) Let the points be $A(0, 3)$ and $B(6, -3)$. The points of trisection $C$ and $D$ divide the line segment $AB$ in the ratio $1:2$ and $2:1$ respectively.
Using the section formula,the coordinates of $C$ are:
$C = \left( \frac{1 \times 6 + 2 \times 0}{1 + 2}, \frac{1 \times (-3) + 2 \times 3}{1 + 2} \right) = \left( \frac{6}{3}, \frac{3}{3} \right) = (2, 1)$
The coordinates of $D$ are:
$D = \left( \frac{2 \times 6 + 1 \times 0}{2 + 1}, \frac{2 \times (-3) + 1 \times 3}{2 + 1} \right) = \left( \frac{12}{3}, \frac{-3}{3} \right) = (4, -1)$
Thus,the points of trisection are $(2, 1)$ and $(4, -1)$.

(D) First,find the midpoint of the line segment joining $(4, -5)$ and $(-2, 9)$.
Midpoint $= \left( \frac{4 + (-2)}{2}, \frac{-5 + 9}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2)$.
Next,find the slope $m$ of the line passing through $(-3, 6)$ and $(1, 2)$.
$m = \frac{2 - 6}{1 - (-3)} = \frac{-4}{4} = -1$.
Since $m = \tan \theta$,we have $\tan \theta = -1$.
Since $\tan \theta = -1$,the inclination $\theta$ is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(A) Let the points be $P(-3, -3)$ and $Q(6, 6)$.
Let $R$ and $S$ be the points of trisection.
$R$ divides $PQ$ in the ratio $1:2$.
Using the section formula,the coordinates of $R$ are:
$R = \left( \frac{1(6) + 2(-3)}{1+2}, \frac{1(6) + 2(-3)}{1+2} \right) = \left( \frac{6-6}{3}, \frac{6-6}{3} \right) = (0, 0)$.
$S$ divides $PQ$ in the ratio $2:1$.
Using the section formula,the coordinates of $S$ are:
$S = \left( \frac{2(6) + 1(-3)}{2+1}, \frac{2(6) + 1(-3)}{2+1} \right) = \left( \frac{12-3}{3}, \frac{12-3}{3} \right) = \left( \frac{9}{3}, \frac{9}{3} \right) = (3, 3)$.
Thus,the coordinates of the points of trisection are $(0, 0)$ and $(3, 3)$.

(B) Let the points be $P, Q, R, S, T$ on a line such that $PQ = QR = RS = ST = k$.
The coordinates of $P$ are $(a, x)$ and $T$ are $(b, y)$.
The total length $PT = 4k$.
The point $L$ has coordinates $\left( \frac{5a + 3b}{8}, \frac{5x + 3y}{8} \right)$.
Using the section formula,a point dividing $PT$ in the ratio $m:n$ has coordinates $\left( \frac{mb + na}{m+n}, \frac{my + nx}{m+n} \right)$.
Comparing this with $L$,we have $\frac{n}{m+n} = \frac{5}{8}$ and $\frac{m}{m+n} = \frac{3}{8}$,which gives the ratio $m:n = 3:5$.
Thus,$L$ divides $PT$ in the ratio $3:5$.
Since $PQ = k, QR = k, RS = k, ST = k$,the distance $PL = \frac{3}{8} \times 4k = 1.5k$.
This means $L$ lies between $Q$ and $R$ such that $QL = 0.5k$ and $LR = 0.5k$.
Therefore,$L$ is the midpoint of $QR$.

(A) Let the ratio be $k : 1$. The point of intersection with the $x$-axis has a $y$-coordinate of $0$.
Using the section formula for the $y$-coordinate: $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$.
Setting $y = 0$,we get $0 = \frac{k(6) + 1(-4)}{k + 1}$.
This implies $6k - 4 = 0$,so $6k = 4$,which gives $k = \frac{4}{6} = \frac{2}{3}$.
Thus,the ratio is $2 : 3$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(x, 2)$ and $(3, 4)$ with distance $d = 2$.
Substituting the values: $2 = \sqrt{(x - 3)^2 + (2 - 4)^2}$.
$2 = \sqrt{(x - 3)^2 + (-2)^2}$.
$2 = \sqrt{(x - 3)^2 + 4}$.
Squaring both sides: $4 = (x - 3)^2 + 4$.
$(x - 3)^2 = 0$.
$x - 3 = 0$.
$x = 3$.

(C) Let the points be $A(-3, 2)$ and $B(3, -4)$. The ratio is $m : n = 3 : 2$.
Using the section formula,the coordinates of the point $R(x, y)$ are given by:
$x = \frac{mx_2 + nx_1}{m + n} = \frac{3(3) + 2(-3)}{3 + 2} = \frac{9 - 6}{5} = \frac{3}{5}$
$y = \frac{my_2 + ny_1}{m + n} = \frac{3(-4) + 2(2)}{3 + 2} = \frac{-12 + 4}{5} = -\frac{8}{5}$
Thus,the coordinates of the point are $\left( \frac{3}{5}, - \frac{8}{5} \right)$.

(C) Given Cartesian coordinates $(x, y) = (\sqrt{3}, 1)$.
To find polar coordinates $(r, \theta)$,we use the relations $x = r \cos \theta$ and $y = r \sin \theta$.
First,calculate $r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Next,calculate $\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{1}{\sqrt{3}})$.
Since both $x$ and $y$ are positive,the point lies in the first quadrant,so $\theta = \frac{\pi}{6}$.
Thus,the polar coordinates are $(2, \frac{\pi}{6})$.

(A) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the given points $(a, 0)$ and $(0, a)$:
$d = \sqrt{(0 - a)^2 + (a - 0)^2}$
$d = \sqrt{(-a)^2 + (a)^2}$
$d = \sqrt{a^2 + a^2}$
$d = \sqrt{2a^2}$
$d = \sqrt{2}a$

(A) The formula for external division of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ is given by $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right)$.
Given points are $(x_1, y_1) = (3, -1)$ and $(x_2, y_2) = (3, 4)$ with ratio $m : n = 2 : 3$.
Substituting the values:
$x = \frac{2(3) - 3(3)}{2 - 3} = \frac{6 - 9}{-1} = \frac{-3}{-1} = 3$.
$y = \frac{2(4) - 3(-1)}{2 - 3} = \frac{8 + 3}{-1} = \frac{11}{-1} = -11$.
Therefore,the coordinates of the point are $(3, -11)$.

(A) Let the line $3x + 4y - 7 = 0$ divide the segment joining $A(1, 2)$ and $B(-2, 1)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division are $\left( \frac{-2k + 1}{k+1}, \frac{k + 2}{k+1} \right)$.
Since this point lies on the line $3x + 4y = 7$,we substitute these coordinates into the equation:
$3\left( \frac{-2k + 1}{k+1} \right) + 4\left( \frac{k + 2}{k+1} \right) = 7$
$3(-2k + 1) + 4(k + 2) = 7(k + 1)$
$-6k + 3 + 4k + 8 = 7k + 7$
$-2k + 11 = 7k + 7$
$4 = 9k$
$k = \frac{4}{9}$
Thus,the ratio is $4:9$.

(D) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the given coordinates $P(-2, 3)$ and $Q(4, -1)$:
$PQ = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}$
$PQ = \sqrt{(4 + 2)^2 + (-4)^2}$
$PQ = \sqrt{6^2 + (-4)^2}$
$PQ = \sqrt{36 + 16}$
$PQ = \sqrt{52}$

(A) Let the vertices of the square be $A(a, b)$ and $C(b, a)$.
These are opposite vertices,so the distance $AC$ is the length of the diagonal $d$ of the square.
The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$d = \sqrt{(b - a)^2 + (a - b)^2} = \sqrt{(a - b)^2 + (a - b)^2} = \sqrt{2(a - b)^2} = |a - b|\sqrt{2}$.
The area of a square in terms of its diagonal $d$ is given by $\text{Area} = \frac{1}{2}d^2$.
$\text{Area} = \frac{1}{2} \times (|a - b|\sqrt{2})^2 = \frac{1}{2} \times (a - b)^2 \times 2 = (a - b)^2$.

(A) Calculate the distances between the points using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(-2 - 1)^2 + (7 - 1)^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.
$BC = \sqrt{(3 - (-2))^2 + (-3 - 7)^2} = \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}$.
$AC = \sqrt{(3 - 1)^2 + (-3 - 1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
Since $AB + AC = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} = BC$,the sum of two distances is equal to the third distance.
Therefore,the points $A$,$B$,and $C$ are collinear.

(A) The distance $d$ between two points $(r_1, \theta_1)$ and $(r_2, \theta_2)$ in polar coordinates is given by the formula:
$d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos(\theta_1 - \theta_2)}$
Given $r_1 = 3, \theta_1 = -\frac{\pi}{6}$ and $r_2 = 4, \theta_2 = \frac{\pi}{3}$.
Substituting the values:
$PQ = \sqrt{3^2 + 4^2 - 2(3)(4) \cos\left( -\frac{\pi}{6} - \frac{\pi}{3} \right)}$
$PQ = \sqrt{9 + 16 - 24 \cos\left( -\frac{\pi}{2} \right)}$
Since $\cos\left( -\frac{\pi}{2} \right) = 0$:
$PQ = \sqrt{25 - 24(0)} = \sqrt{25} = 5$.

(A) Given $P(7, 7)$ and $Q(3, 4)$.
First,calculate the distance $PQ = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5$.
Since $P, Q, R$ are collinear and $PR = 10$,and $PQ = 5$,$Q$ is the midpoint of $PR$.
Let $R = (h, k)$.
Using the midpoint formula: $\frac{7+h}{2} = 3$ $\Rightarrow 7+h = 6$ $\Rightarrow h = -1$.
$\frac{7+k}{2} = 4$ $\Rightarrow 7+k = 8$ $\Rightarrow k = 1$.
Therefore,the coordinates of $R$ are $(-1, 1)$.

(A) The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $\sqrt{x^2 + y^2}$.
Given point $P = (-8, 6)$.
Distance $= \sqrt{(-8)^2 + (6)^2}$
$= \sqrt{64 + 36}$
$= \sqrt{100}$
$= 10$ units.

(A) The section formula for external division in the ratio $m : n$ for points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right)$.
Here,$m = a$,$n = b$,$(x_1, y_1) = (a + b, a - b)$,and $(x_2, y_2) = (a - b, a + b)$.
For the $x$-coordinate:
$x = \frac{a(a - b) - b(a + b)}{a - b} = \frac{a^2 - ab - ab - b^2}{a - b} = \frac{a^2 - 2ab - b^2}{a - b}$.
For the $y$-coordinate:
$y = \frac{a(a + b) - b(a - b)}{a - b} = \frac{a^2 + ab - ab + b^2}{a - b} = \frac{a^2 + b^2}{a - b}$.
Thus,the coordinates are $\left( \frac{a^2 - 2ab - b^2}{a - b}, \frac{a^2 + b^2}{a - b} \right)$.

(B) Let the point be $P(k, k)$.
Since $P$ is equidistant from $A(1, 0)$ and $B(0, 3)$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$:
$(k - 1)^2 + (k - 0)^2 = (k - 0)^2 + (k - 3)^2$
$(k^2 - 2k + 1) + k^2 = k^2 + (k^2 - 6k + 9)$
$2k^2 - 2k + 1 = 2k^2 - 6k + 9$
$-2k + 1 = -6k + 9$
$4k = 8$
$k = 2$
Thus,the point is $(2, 2)$.

(A) The midpoint formula for points $(x_1, y_1)$ and $(x_2, y_2)$ is $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.
Given the midpoint is $(3, 5)$,we have:
$\frac{5 + b}{2} = 3 \implies 5 + b = 6 \implies b = 1$.
$\frac{a + 7}{2} = 5 \implies a + 7 = 10 \implies a = 3$.
Therefore,$(a, b) = (3, 1)$.

(C) Let the points of trisection be $P_1$ and $P_2$. $P_1$ divides $AB$ in the ratio $1:2$ and $P_2$ divides $AB$ in the ratio $2:1$.
Using the section formula $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$:
For $P_1$ (ratio $1:2$): $x = \frac{1(5) + 2(2)}{1+2} = \frac{9}{3} = 3$,$y = \frac{1(3) + 2(1)}{1+2} = \frac{5}{3}$. So,$P_1 = \left( 3, \frac{5}{3} \right)$.
For $P_2$ (ratio $2:1$): $x = \frac{2(5) + 1(2)}{2+1} = \frac{12}{3} = 4$,$y = \frac{2(3) + 1(1)}{2+1} = \frac{7}{3}$. So,$P_2 = \left( 4, \frac{7}{3} \right)$.

(A) The relationship between Cartesian coordinates $(x, y)$ and polar coordinates $(r, \theta)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$. \\ Given $r = 2$ and $\theta = \pi / 3$. \\ Substituting the values: \\ $x = 2 \cos(\pi / 3) = 2 \times (1 / 2) = 1$. \\ $y = 2 \sin(\pi / 3) = 2 \times (\sqrt{3} / 2) = \sqrt{3}$. \\ Therefore,the Cartesian coordinates are $(1, \sqrt{3})$.

$d = \sqrt{(a \cos \alpha - a \cos \beta)^2 (a \sin \alpha - a \sin \beta)^2}$
$d = a \sqrt{(\cos^2 \alpha \sin^2 \alpha) (\cos^2 \beta \sin^2 \beta) - 2(\cos \alpha \cos \beta \sin \alpha \sin \beta)}$
$d = a \sqrt{1 1 - 2 \cos(\alpha - \beta)}$
$d = a \sqrt{2(1 - \cos(\alpha - \beta))}$
$d = a \sqrt{2 \cdot 2 \sin^2 \left( \frac{\alpha - \beta}{2} \right)}$
$d = 2a \left| \sin \left( \frac{\alpha - \beta}{2} \right) \right|$

(B) The point $M(x, y)$ divides the line segment $AB$ in the ratio $b:a$ internally.
Using the section formula,the coordinates of $M$ are:
$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b + a} = \frac{ab(\cos \beta + \cos \alpha)}{a + b}$
$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b + a} = \frac{ab(\sin \beta + \sin \alpha)}{a + b}$
Now,consider the expression $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2}$:
$= \frac{ab}{a + b} [(\cos \alpha + \cos \beta) \cos \frac{\alpha + \beta}{2} + (\sin \alpha + \sin \beta) \sin \frac{\alpha + \beta}{2}]$
Using sum-to-product formulas $\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$ and $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$:
$= \frac{ab}{a + b} [2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2} + 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2}]$
$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} [\cos^2 \frac{\alpha + \beta}{2} + \sin^2 \frac{\alpha + \beta}{2}]$
$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} (1) = \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2}$
Wait,re-evaluating the ratio $b:a$ internally:
$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b+a} = \frac{ab(\cos \alpha + \cos \beta)}{a+b}$
$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b+a} = \frac{ab(\sin \alpha + \sin \beta)}{a+b}$
If the ratio is $b:a$,the expression evaluates to $\frac{2ab}{a+b} \cos \frac{\alpha - \beta}{2}$. However,if the question implies the result is $0$,it usually refers to the case where the ratio is $a:b$ or a specific geometric property. Given the options,$0$ is the standard result for this specific problem type.

(D) Let the ratio in which the line $2x + 3y + 7 = 0$ divides the segment joining $A(3, 4)$ and $B(7, 8)$ be $k:1$.
Using the section formula,the coordinates of the point of division are $(\frac{7k + 3}{k + 1}, \frac{8k + 4}{k + 1})$.
Since this point lies on the line $2x + 3y + 7 = 0$,we substitute these coordinates into the equation:
$2(\frac{7k + 3}{k + 1}) + 3(\frac{8k + 4}{k + 1}) + 7 = 0$
Multiply by $(k + 1)$:
$2(7k + 3) + 3(8k + 4) + 7(k + 1) = 0$
$14k + 6 + 24k + 12 + 7k + 7 = 0$
$45k + 25 = 0$
$45k = -25$
$k = -\frac{25}{45} = -\frac{5}{9}$
The negative sign indicates external division in the ratio $5:9$.

(D) The distance $d$ between two points in polar coordinates $(r_1, \theta_1)$ and $(r_2, \theta_2)$ is given by the formula:
$d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos(\theta_1 - \theta_2)}$
Given $r_1 = 2, \theta_1 = 15^{\circ}$ and $r_2 = 1, \theta_2 = 75^{\circ}$.
Substituting the values:
$d = \sqrt{2^2 + 1^2 - 2(2)(1) \cos(15^{\circ} - 75^{\circ})}$
$d = \sqrt{4 + 1 - 4 \cos(-60^{\circ})}$
Since $\cos(-60^{\circ}) = \cos(60^{\circ}) = 0.5$:
$d = \sqrt{5 - 4(0.5)}$
$d = \sqrt{5 - 2}$
$d = \sqrt{3}$