Transformation, Pedal points Questions in English

(B) Step $1$: Reflection of the point $(x, y)$ about the line $y = x$ results in the point $(y, x)$.
Applying this to the point $(4, 1)$,we get $(1, 4)$.
Step $2$: Translation of the point $(x, y)$ through a distance $d$ along the positive $x$-axis results in the point $(x + d, y)$.
Applying this to the point $(1, 4)$ with $d = 2$,we get $(1 + 2, 4) = (3, 4)$.
Thus,the final coordinates of the point are $(3, 4)$.

(C) Step $1$: Reflection of $(4, 1)$ about $y = x$ gives $(1, 4)$.
Step $2$: Translation by $2$ units along the positive $x$-axis: $(1 + 2, 4) = (3, 4)$.
Step $3$: Rotation of $(3, 4)$ by $\theta = \pi/4$ counter-clockwise about the origin.
The new coordinates $(x', y')$ are given by:
$x' = x \cos \theta - y \sin \theta = 3 \cos(\pi/4) - 4 \sin(\pi/4) = 3(\frac{1}{\sqrt{2}}) - 4(\frac{1}{\sqrt{2}}) = -\frac{1}{\sqrt{2}}$
$y' = x \sin \theta + y \cos \theta = 3 \sin(\pi/4) + 4 \cos(\pi/4) = 3(\frac{1}{\sqrt{2}}) + 4(\frac{1}{\sqrt{2}}) = \frac{7}{\sqrt{2}}$
Thus,the final position is $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$.

(C) Let $ABC$ be the triangle and $D(20, 25)$,$E(8, 16)$,$F(8, 9)$ be the feet of the perpendiculars from $A, B, C$ respectively. Then $DEF$ is the pedal triangle of $\Delta ABC$.
From geometry,we know that the orthocentre of $\Delta ABC$ is the incentre of its pedal triangle $DEF$.
Let the coordinates of the orthocentre be $O(h, k)$.
First,calculate the side lengths of $\Delta DEF$:
$DE = \sqrt{(20-8)^2 + (25-16)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$
$EF = \sqrt{(8-8)^2 + (16-9)^2} = \sqrt{0^2 + 7^2} = 7$
$FD = \sqrt{(20-8)^2 + (25-9)^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$
Using the incentre formula $I = \frac{aA + bB + cC}{a+b+c}$,where $a, b, c$ are side lengths opposite to vertices $D, E, F$ respectively:
$h = \frac{EF \cdot x_D + FD \cdot x_E + DE \cdot x_F}{EF + FD + DE} = \frac{7(20) + 20(8) + 15(8)}{7 + 20 + 15} = \frac{140 + 160 + 120}{42} = \frac{420}{42} = 10$
$k = \frac{EF \cdot y_D + FD \cdot y_E + DE \cdot y_F}{EF + FD + DE} = \frac{7(25) + 20(16) + 15(9)}{7 + 20 + 15} = \frac{175 + 320 + 135}{42} = \frac{630}{42} = 15$
Thus,the orthocentre is $(10, 15)$.

(D) The line $L$ is given by $x - y = 4$,which has a slope $m = 1$.
Since the point $P(2, 1)$ is translated parallel to $L$,the new point $Q(x, y)$ lies on a line parallel to $L$.
The slope of the line passing through $P$ and $Q$ is $1$.
The distance between $P(2, 1)$ and $Q(x, y)$ is $2\sqrt{3}$.
Using the parametric form of a line,the coordinates of $Q$ are $(2 \pm 2\sqrt{3} \cos \theta, 1 \pm 2\sqrt{3} \sin \theta)$,where $\tan \theta = 1$,so $\cos \theta = \sin \theta = \frac{1}{\sqrt{2}}$.
$Q = (2 \pm 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}, 1 \pm 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}) = (2 \pm \sqrt{6}, 1 \pm \sqrt{6})$.
Since $Q$ lies in the third quadrant,both coordinates must be negative.
Choosing the negative sign: $Q = (2 - \sqrt{6}, 1 - \sqrt{6})$.
The line passing through $Q$ and perpendicular to $L$ has a slope $m' = -1$.
The equation is $y - (1 - \sqrt{6}) = -1(x - (2 - \sqrt{6}))$.
$y - 1 + \sqrt{6} = -x + 2 - \sqrt{6}$.
$x + y = 3 - 2\sqrt{6}$.

(A) The reflection of point $P(a, b)$ about the line $y=x$ gives the point $(b, a)$.
Translating this point by $2$ units along the positive $x$-axis results in the point $(b+2, a)$.
Applying a rotation of $\frac{\pi}{4}$ anti-clockwise about the origin,the new coordinates $(x', y')$ are given by:
$x' = (b+2)\cos\frac{\pi}{4} - a\sin\frac{\pi}{4} = \frac{b+2-a}{\sqrt{2}}$
$y' = (b+2)\sin\frac{\pi}{4} + a\cos\frac{\pi}{4} = \frac{b+2+a}{\sqrt{2}}$
Given the final position is $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$,we have:
$\frac{b+2-a}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$ $\Rightarrow b-a+2 = -1$ $\Rightarrow b-a = -3$ (Equation $1$)
$\frac{b+2+a}{\sqrt{2}} = \frac{7}{\sqrt{2}}$ $\Rightarrow b+a+2 = 7$ $\Rightarrow b+a = 5$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$2b = 2 \Rightarrow b = 1$
Substituting $b=1$ into Equation $2$:
$1+a = 5 \Rightarrow a = 4$
Therefore,$2a+b = 2(4)+1 = 9$.

(A) Let the coordinates of $A$ be $(x_1, y_1) = (2, 0)$ and $B$ be $(x_2, y_2) = (3, 1)$.
The length of $AB$ is $r = \sqrt{(3-2)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\theta$ that $AB$ makes with the positive $x$-axis is given by $\tan \theta = \frac{1-0}{3-2} = 1$,so $\theta = 45^{\circ}$.
When the line is rotated by $45^{\circ}$ in the anti-clockwise direction about $A$,the new angle $\theta'$ becomes $\theta + 45^{\circ} = 45^{\circ} + 45^{\circ} = 90^{\circ}$.
Let the new coordinates of $B$ be $(x', y')$.
Using the rotation formula,$x' = x_1 + r \cos \theta'$ and $y' = y_1 + r \sin \theta'$.
$x' = 2 + \sqrt{2} \cos 90^{\circ} = 2 + \sqrt{2}(0) = 2$.
$y' = 0 + \sqrt{2} \sin 90^{\circ} = 0 + \sqrt{2}(1) = \sqrt{2}$.
Thus,the new coordinates are $(2, \sqrt{2})$.

(A) Let $A = (x_1, y_1) = (2, 0)$ and $B = (x_2, y_2) = (6, 4)$.
Vector $\vec{AB} = (6-2, 4-0) = (4, 4)$.
The length of $AB$ is $r = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.
The angle of $\vec{AB}$ with the positive $x$-axis is $\theta = \tan^{-1}(\frac{4}{4}) = 45^{\circ}$.
Rotating the vector by $45^{\circ}$ in the negative (clockwise) direction gives a new angle $\theta' = 45^{\circ} - 45^{\circ} = 0^{\circ}$.
The new coordinates $(x', y')$ of $B$ are given by $x' = x_1 + r \cos(\theta')$ and $y' = y_1 + r \sin(\theta')$.
$x' = 2 + 4\sqrt{2} \cos(0^{\circ}) = 2 + 4\sqrt{2}(1) = 2 + 4\sqrt{2}$.
$y' = 0 + 4\sqrt{2} \sin(0^{\circ}) = 0 + 4\sqrt{2}(0) = 0$.
Thus,the new coordinates are $(2 + 4\sqrt{2}, 0)$.

(C) Step $1$: Reflection of $(4,1)$ about the line $y=x$ gives $(1,4)$.
Step $2$: Translation of $(1,4)$ by $2$ units in the positive $X$-direction gives $(1+2, 4) = (3,4)$.
Step $3$: Rotation of $(x,y) = (3,4)$ by an angle $\theta = \frac{\pi}{4}$ anticlockwise about the origin is given by the transformation:
$x' = x \cos \theta - y \sin \theta$
$y' = x \sin \theta + y \cos \theta$
Substituting the values:
$x' = 3 \cos \frac{\pi}{4} - 4 \sin \frac{\pi}{4} = 3 \left(\frac{1}{\sqrt{2}}\right) - 4 \left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}}$
$y' = 3 \sin \frac{\pi}{4} + 4 \cos \frac{\pi}{4} = 3 \left(\frac{1}{\sqrt{2}}\right) + 4 \left(\frac{1}{\sqrt{2}}\right) = \frac{7}{\sqrt{2}}$
Thus,the final position is $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

(C) Given point is $(3, 2)$.
$(i)$ Reflection of point $(3, 2)$ about the line $y=x$ gives $(2, 3)$.
(ii) Translation of point $(2, 3)$ by $1$ unit in the positive $x$-direction gives $(2+1, 3) = (3, 3)$.
(iii) Rotation of point $(x, y) = (3, 3)$ by an angle $\theta = \frac{\pi}{4}$ about the origin in the anti-clockwise direction is given by the transformation:
$X = x \cos \theta - y \sin \theta = 3 \cos(\frac{\pi}{4}) - 3 \sin(\frac{\pi}{4}) = 3(\frac{1}{\sqrt{2}}) - 3(\frac{1}{\sqrt{2}}) = 0$
$Y = x \sin \theta + y \cos \theta = 3 \sin(\frac{\pi}{4}) + 3 \cos(\frac{\pi}{4}) = 3(\frac{1}{\sqrt{2}}) + 3(\frac{1}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2} = \sqrt{18}$
Thus,the final position is $(0, \sqrt{18})$.