Questions related to geometrical conditions Questions in English

(A) Let the points be $A(-a, -b)$,$B(0, 0)$,$C(a, b)$,and $D(a^2, ab)$.
To check if the points are collinear,we can check the slopes between consecutive points.
Slope of $AB = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a}$.
Slope of $BC = \frac{b - 0}{a - 0} = \frac{b}{a}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.
Now,check the slope of $CD = \frac{ab - b}{a^2 - a} = \frac{b(a - 1)}{a(a - 1)} = \frac{b}{a}$ (assuming $a \neq 1, 0$).
Since the slopes of $AB, BC,$ and $CD$ are all equal to $\frac{b}{a}$,all four points lie on the same line.
Therefore,the points are collinear.

(C) Let the vertices be $A(-2, 2), B(-2, -1), C(3, -1),$ and $D(3, 2)$.
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(-2 - (-2))^2 + (-1 - 2)^2} = \sqrt{0^2 + (-3)^2} = 3$
$BC = \sqrt{(3 - (-2))^2 + (-1 - (-1))^2} = \sqrt{5^2 + 0^2} = 5$
$CD = \sqrt{(3 - 3)^2 + (2 - (-1))^2} = \sqrt{0^2 + 3^2} = 3$
$DA = \sqrt{(-2 - 3)^2 + (2 - 2)^2} = \sqrt{(-5)^2 + 0^2} = 5$
Since opposite sides are equal ($AB = CD = 3$ and $BC = DA = 5$) and adjacent sides are perpendicular (as sides are parallel to the coordinate axes),the figure is a rectangle.

(A) The diagonals of a square bisect each other at $90^{\circ}$ at the origin. Let the vertices be on the axes. The distance from the origin to each vertex is half the length of the diagonal.
Since the side length is $a$,the diagonal length $d$ is given by $d = a\sqrt{2}$.
The distance from the origin to each vertex is $\frac{d}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Thus,the vertices of the square are $\left(\frac{a}{\sqrt{2}}, 0\right)$,$\left(-\frac{a}{\sqrt{2}}, 0\right)$,$\left(0, \frac{a}{\sqrt{2}}\right)$,and $\left(0, -\frac{a}{\sqrt{2}}\right)$.
Comparing these with the given options,the point $(a\sqrt{2}, 0)$ is not a vertex of the square.

(D) Let the coordinates of point $C$ be $(h, k)$.
Given $\angle A - \angle B = \theta$,so $\tan(A - B) = \tan \theta$ .....$(i)$
In the right-angled triangle $CDA$,where $D$ is the projection of $C$ on the $x$-axis,$\tan A = \frac{k}{a - h}$.
In the right-angled triangle $CDB$,$\tan B = \frac{k}{h - (-a)} = \frac{k}{h + a}$.
From $(i)$,we have $\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan \theta$.
Substituting the values of $\tan A$ and $\tan B$:
$\frac{\frac{k}{a - h} - \frac{k}{a + h}}{1 + \frac{k^2}{a^2 - h^2}} = \tan \theta$
$\frac{k(a + h) - k(a - h)}{a^2 - h^2 + k^2} = \tan \theta$
$\frac{2kh}{a^2 - h^2 + k^2} = \tan \theta$
$2kh = (a^2 - h^2 + k^2) \tan \theta$
$2kh \cot \theta = a^2 - h^2 + k^2$
$h^2 - k^2 + 2hk \cot \theta = a^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} - {y^2} + 2xy \cot \theta = {a^2}$.

(B) The Cartesian coordinates $(x, y)$ of a point with polar coordinates $(r, \theta)$ are given by $x = r \cos \theta$ and $y = r \sin \theta$.
Given $r = \sqrt{2}$ and $\theta = -\frac{3\pi}{4}$.
$x = \sqrt{2} \cos\left( -\frac{3\pi}{4} \right) = \sqrt{2} \cos\left( \frac{3\pi}{4} \right) = \sqrt{2} \left( -\frac{1}{\sqrt{2}} \right) = -1$.
$y = \sqrt{2} \sin\left( -\frac{3\pi}{4} \right) = -\sqrt{2} \sin\left( \frac{3\pi}{4} \right) = -\sqrt{2} \left( \frac{1}{\sqrt{2}} \right) = -1$.
Thus,the Cartesian coordinates are $(-1, -1)$.

(C) The initial position of the insect is $A(3, 2)$.
Moving $4 \ units$ towards the west (negative $x$-direction) changes the $x$-coordinate: $3 - 4 = -1$. The position becomes $(-1, 2)$.
Moving $3 \ units$ towards the south (negative $y$-direction) changes the $y$-coordinate: $2 - 3 = -1$. The final position $B$ is $(-1, -1)$.
To find the polar coordinates $(r, \theta)$ for $B(-1, -1)$:
$r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Since the point $(-1, -1)$ lies in the third quadrant,the angle $\theta$ is given by $\theta = -\pi + \tan^{-1}\left(\frac{-1}{-1}\right) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.
Thus,the polar coordinates are $\left( \sqrt{2}, -\frac{3\pi}{4} \right)$.

(B) Let the vertices of the square be $O(0, 0)$,$A$,$B$,and $C$. The side $OA$ makes an angle of $30^o$ with the positive $x-$axis. Since the side length is $2$,the coordinates of $A$ are $(2 \cos 30^o, 2 \sin 30^o) = (2 \cdot \frac{\sqrt{3}}{2}, 2 \cdot \frac{1}{2}) = (\sqrt{3}, 1)$.
The side $OC$ is perpendicular to $OA$ and lies at an angle of $30^o + 90^o = 120^o$ with the positive $x-$axis. The coordinates of $C$ are $(2 \cos 120^o, 2 \sin 120^o) = (2 \cdot -\frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}) = (-1, \sqrt{3})$.
The vertex $B$ is the sum of vectors $\vec{OA}$ and $\vec{OC}$,so $B = (\sqrt{3} - 1, 1 + \sqrt{3})$.
The $x-$coordinates of the vertices are $0$,$\sqrt{3}$,$-1$,and $\sqrt{3} - 1$.
The sum of the $x-$coordinates is $0 + \sqrt{3} - 1 + \sqrt{3} - 1 = 2\sqrt{3} - 2$.

(B) Let the side length of the square $ABCD$ be $x$. The area of the square is $x^2$.
Let $M$ be the intersection of the diagonals $AC$ and $BD$. Since $ABCD$ is a square,the diagonals bisect each other at right angles,so $AM = MC = BM = MD = \frac{x}{\sqrt{2}}$.
Since $E, A, C$ are collinear,$EA$ lies on the diagonal $AC$. In $\triangle EBD$,$EB = ED = \sqrt{130}$ and $BD$ is the base. The altitude from $E$ to $BD$ is $EM$. Since $\triangle EBD$ is isosceles,$EM \perp BD$.
In $\triangle EBM$,by the Pythagorean theorem,$EM^2 + BM^2 = EB^2$.
$EM^2 + (\frac{x}{\sqrt{2}})^2 = 130$ $\Rightarrow EM^2 = 130 - \frac{x^2}{2}$ $\Rightarrow EM = \sqrt{130 - \frac{x^2}{2}}$.
The area of $\triangle EAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times EA \times \sin(\angle EAB)$. Since $E, A, C$ are collinear,$\angle EAB = 135^\circ$. Thus,$\text{Area} = \frac{1}{2} x \cdot EA \cdot \sin(135^\circ) = \frac{1}{2} x \cdot EA \cdot \frac{1}{\sqrt{2}}$.
Alternatively,using the coordinates with $A$ at $(0,0)$,$B$ at $(x,0)$,$D$ at $(0,x)$,and $C$ at $(x,x)$,the line $AC$ is $y=x$. Point $E$ lies on $y=x$ with $x_E < 0$,so $E = (-a, -a)$.
$EB^2 = (x+a)^2 + a^2 = 130$ and $ED^2 = a^2 + (x+a)^2 = 130$. Area of $\triangle EAB = \frac{1}{2} |x_A(y_B-y_E) + x_B(y_E-y_A) + x_E(y_A-y_B)| = \frac{1}{2} |0 + x(-a-0) + (-a)(0-0)| = \frac{ax}{2}$.
Given $\frac{ax}{2} = x^2 \Rightarrow a = 2x$.
Substitute $a=2x$ into $EB^2 = (x+2x)^2 + (2x)^2 = 130$ $\Rightarrow 9x^2 + 4x^2 = 130$ $\Rightarrow 13x^2 = 130$ $\Rightarrow x^2 = 10$.
The area of the square is $x^2 = 10$.

(C) Given Cartesian coordinates $(x, y) = \left(-\frac{5 \sqrt{3}}{2}, \frac{5}{2}\right)$.
$r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{5 \sqrt{3}}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{75}{4} + \frac{25}{4}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5$.
Since the point lies in the second quadrant $(x < 0, y > 0)$,the polar angle $\theta$ is given by $\theta = \pi - \tan^{-1}\left|\frac{y}{x}\right|$.
$\theta = \pi - \tan^{-1}\left|\frac{5/2}{-5\sqrt{3}/2}\right| = \pi - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.
Thus,the polar coordinates are $(r, \theta) = \left(5, \frac{5 \pi}{6}\right)$.

(D) Given polar coordinates are $P(r, \theta) = \left(\frac{1}{2}, 120^{\circ}\right)$,where $r = \frac{1}{2}$ and $\theta = 120^{\circ}$.
We use the conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$.
For $x$: $x = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} \left(-\frac{1}{2}\right) = -\frac{1}{4}$.
For $y$: $y = \frac{1}{2} \sin 120^{\circ} = \frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}$.
Therefore,the Cartesian coordinates are $\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$.

(B) Given Cartesian coordinates $(x, y) = (-\sqrt{2}, \sqrt{2})$.
To find polar coordinates $(r, \theta)$:
$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.
Since $x = r \cos \theta$ and $y = r \sin \theta$,we have:
$\cos \theta = \frac{-\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,the point lies in the second quadrant.
$\theta = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.
Thus,the polar coordinates are $\left(2, \frac{3 \pi}{4}\right)$.

(C) The given equation is $4x^2 + 9y^2 - 8x + 36y + 4 = 0$.
To eliminate the first-degree terms,we complete the square for $x$ and $y$ terms:
$4(x^2 - 2x) + 9(y^2 + 4y) + 4 = 0$
$4(x^2 - 2x + 1 - 1) + 9(y^2 + 4y + 4 - 4) + 4 = 0$
$4(x - 1)^2 - 4 + 9(y + 2)^2 - 36 + 4 = 0$
$4(x - 1)^2 + 9(y + 2)^2 = 36$
Let the new coordinates be $X = x - 1$ and $Y = y + 2$.
This implies $x = X + 1$ and $y = Y - 2$.
Thus,the origin must be shifted to the point $(1, -2)$ to eliminate the $x$ and $y$ terms.

(A) Given equation is $4x^2 + 9y^2 - 8x + 36y + 4 = 0$.
To eliminate the $x$ and $y$ terms,we shift the origin to $(h, k)$.
Let $x = X + h$ and $y = Y + k$.
Substituting these into the equation: $4(X+h)^2 + 9(Y+k)^2 - 8(X+h) + 36(Y+k) + 4 = 0$.
Expanding the terms: $4(X^2 + 2hX + h^2) + 9(Y^2 + 2kY + k^2) - 8X - 8h + 36Y + 36k + 4 = 0$.
Grouping the linear terms: $(8h - 8)X + (18k + 36)Y + (4h^2 + 9k^2 - 8h + 36k + 4) = 0$.
For the $X$ and $Y$ terms to be eliminated,their coefficients must be zero:
$8h - 8 = 0 \implies h = 1$.
$18k + 36 = 0 \implies k = -2$.
Thus,the origin is shifted to $(1, -2)$.

(D) Given,square $ABCD$ with vertices $A(0,0), B(2,0), C(2,2), D(0,2)$.
Applying $f_1(x, y) \longrightarrow (y, x)$:
$A(0,0)$ $\longrightarrow A'(0,0), B(2,0)$ $\longrightarrow B'(0,2), C(2,2)$ $\longrightarrow C'(2,2), D(0,2)$ $\longrightarrow D'(2,0)$.
Applying $f_2(x, y) \longrightarrow (x+3y, y)$:
$A'(0,0)$ $\longrightarrow A''(0,0), B'(0,2)$ $\longrightarrow B''(6,2), C'(2,2)$ $\longrightarrow C''(8,2), D'(2,0)$ $\longrightarrow D''(2,0)$.
Applying $f_3(x, y) \longrightarrow \left(\frac{x-y}{2}, \frac{x+y}{2}\right)$:
$A''(0,0)$ $\longrightarrow A'''(0,0), B''(6,2)$ $\longrightarrow B'''(2,4), C''(8,2)$ $\longrightarrow C'''(3,5), D''(2,0)$ $\longrightarrow D'''(1,1)$.
Final vertices: $A(0,0), B(2,4), C(3,5), D(1,1)$.
Calculating side lengths:
$AB = \sqrt{(2-0)^2 + (4-0)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
$BC = \sqrt{(3-2)^2 + (5-4)^2} = \sqrt{1+1} = \sqrt{2}$.
$CD = \sqrt{(1-3)^2 + (1-5)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
$DA = \sqrt{(0-1)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Since opposite sides are equal ($AB=CD$ and $BC=DA$) and adjacent sides are not equal,the figure is a parallelogram.