Transformation - class-X

Slope of the line $\displaystyle AB=\frac { 0-1 }{ 2-3 } =1$
$\therefore \angle BAX={ 45 }^{ o }$
Given $\angle B'AB={ 15 }^{ o }\Rightarrow \angle B'AX={ 30 }^{ o }$
Therefore slope of the line $\displaystyle AB'=\tan { { 30 }^{ o } } =\frac { 1 }{ \sqrt { 3 }  } $
Now line $AB'$ makes an angle of ${ 30 }^{ o }$ with positive direction of $x$-axis and 
$AB'=AB=\sqrt { { \left( 3-2 \right)  }^{ 2 }+{ \left( 1-0 \right)  }^{ 2 } } =\sqrt { 2 } $
Therefore coordinates are $\displaystyle \left( 2+\sqrt { 2 } \cos { { 30 }^{ o } } ,0+\sqrt { 2 } \sin { { 30 }^{ o } }  \right) =\left( \frac { 2\sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2 }  } ,\frac { 1 }{ \sqrt { 2 }  }  \right) $

To eliminate the linear terms in the equation x^2 + y^2 + 2x - 4y + 6 = 0, we shift the origin to (alpha, beta) by substituting x with x + alpha and y with y + beta. The linear term for x is 2(x + alpha) + 2(x + alpha) derivative terms, which simplifies to setting the coefficient of x to zero: 2alpha + 2 = 0, giving alpha = -1. For y, setting the coefficient to zero gives -4 - 2(2) or similar derivative method yielding beta = 2. Thus, the shift is (-1, 2).

$ y = x-3$ $\Rightarrow m=1$ ; $tan \theta =1$
By parametrization we have $x=2 \pm r cos\theta $
$y=1 \pm r sin\theta $

Shifting the origin to $(2 , 2)$ transforms the coordinates to
$(x , y)$ to $(x - 2, y - 2)$.

$\therefore$ distance between two points is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$=\sqrt{((x _{1}-2)-(x _{2}-2))^{2}+((y _{1}-2)-(y _{2}-2))}^{2}$

$=\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})}^{2}$

$\therefore$ distance is unaltered $ = 2$ units.

shifted by $\sqrt{2}$ units $\Rightarrow r=\sqrt{2}$

for the line is tan $\theta =1$

$\therefore$ by parametric equations, we have

$x=x^{1}+r\ cos \theta$ ;  $y=y^{1}+r\ sin \theta$

$\Rightarrow x=4+\sqrt{2}\frac{1}{\sqrt{2}}$ ; $y=3+\sqrt{2}\frac{1}{\sqrt{2}}$

$x=5$  ;  $y=4$

Since the line has intecepts a and b on the coordinate axes, therefore its equation is 
$\cfrac { x }{ a } +\cfrac { y }{ b } =1$
When the axes are rotated, its equation with respect to the new axes and the same origin will become
$\cfrac { x }{ p } +\cfrac { y }{ q } =1$
In both the cases, the length of the perpendicular from the origin to the line will be same
Therefore
$\cfrac { 1 }{ \sqrt { \cfrac { 1 }{ { a }^{ 2 } } +\cfrac { 1 }{ { b }^{ 2 } }  }  } =\cfrac { 1 }{ \sqrt { \cfrac { 1 }{ { p }^{ 2 } } +\cfrac { 1 }{ { q }^{ 2 } }  }  } \ \Rightarrow \cfrac { 1 }{ { a }^{ 2 } } +\cfrac { 1 }{ { b }^{ 2 } } =\cfrac { 1 }{ { p }^{ 2 } } +\cfrac { 1 }{ { q }^{ 2 } } $

The given equation is 
$(a-b) (x^2 + y^2) - 2abx = 0$             ........... (i)
The origin is shifted to (ab/(a-b), 0). Any point (x, y) on the curve (i) must be replaced with a new point (X, Y) with reference to new axes, such that
$\displaystyle x = X + \frac{ab}{a-b}  ,    y = Y +0$
substituting these in (i), we get
$(a-b) \displaystyle \left [ \left (X + \frac{ab}{a-b} \right )^2 + y^2 \right ] - 2ab \left [ X+ \frac{ab}{a-b} \right ] = 0$
$\Rightarrow (a-b) \displaystyle \left [ X^2 + \frac{a^2 b^2}{(a-b)^2} + Y^2 + \frac{2abX}{a-b} \right ] - 2abX - \frac{2a^2 b^2}{a-b} = 0$
$\Rightarrow (a-b) (X^2 + Y^2) = \displaystyle \frac{a^2 b^2}{a-b}$
$\Rightarrow (a-b)^2 (X^2+Y^2) = a^2 b^2$

By substituting the new axes X = 2x + y + 2 and Y = x - 2y + 1 into the equation, the expression simplifies directly to 4Y^2 + 9X^2 = 25. The question asks for the equation in terms of X and Y.

Given equation is $x+y=1$
We know that 
$\displaystyle x=X\cos\theta-Y\sin\theta$
$y=X\sin\theta+Y\cos\theta$
$\Rightarrow (\cos\theta+\sin\theta)X+(\cos\theta-\sin\theta)Y=1$  
$\displaystyle \Rightarrow \frac{(\sqrt{3}+1)}{2}X+ \frac{(1-\sqrt{3})}{2}Y=1$       .....(i)
According to problem, we have

The line has direction vector (-1, 3, 1) and passes through (1, 2, 4). The reflection of a line in a plane involves reflecting the direction vector and a point on the line; the normal to the plane is (1, 1, 1). Calculating the reflection of the direction vector across the plane normal yields the new direction vector (-3, -1, 1).

To find the image of a line in another line, find the intersection point and reflect a point from the first line. The intersection of x-y-1=0 and 2x-3y+1=0 is (2, 1). Reflecting a point like (1, 0) from the first line across the second line gives the new line equation.

The image of (1, 2) in y=x is (2, 1). The image of (2, 1) in y=0 is (2, -1). Thus, alpha=2 and beta=-1.

The incident ray x+sqrt(3)y=5 hits the x-axis at (5, 0) with slope -1/sqrt(3), which is -30 degrees. Refraction turns it 30 degrees away from the x-axis, making the new slope tan(30) = 1/sqrt(3). The line passes through (5, 0) with this slope.

The reflection B of A(a, 5) across 4x-3y=0 is found using the reflection formula. The area of triangle ABC (where C is the origin or a fixed point) is calculated using the coordinates of A, B, and the intersection point.

The family of lines (2x-3y+4) + k(x-2y+3) = 0 passes through a fixed point (intersection of the two lines). The locus of the image of a point reflected across a family of lines passing through a fixed point is a circle centered at that fixed point.

Reflecting a point (x, y) across the y-axis results in (-x, y). Reflecting (-5, 4) across the y-axis gives (5, 4).

The line equation can be rewritten as (2x+3y-5)cos(theta) + (3x-5y+2)sin(theta) = 0. This passes through the intersection of 2x+3y=5 and 3x-5y=-2. Solving this gives P(1, 1). Reflecting P across 4x+6y=23 gives Q, then calculate distance to origin.

The image of point A(1, 2) across the line y = x swaps its coordinates, giving point B(2, 1). Next, the image of B(2, 1) across the line y = 0 (the x-axis) reflects the y-coordinate, yielding (2, -1). Therefore, a = 2 and beta = -1.

The reflection of (x1, y1, z1) in ax+by+cz+d=0 is given by (x-x1)/a = (y-y1)/b = (z-z1)/c = -2(ax1+by1+cz1+d)/(a^2+b^2+c^2). Plugging in the values gives the point (26/7, 15/7, 17/7).

Given transformation is 
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right )$
So, the point $(-2,-7)$ $\rightarrow (-2-2(-7),-3(-2)-7)$
So, the image of the point $(-2,-7)$ under the given transformation is $ (12,-1)$

Let points are $A(-2,6)$ & $B(4,2)$

In ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }=0 $ replacing $y$ by $-y$
we get, ${ ax }^{ 2 }+2hx\left( -y \right) +{ by }^{ 2 }=0 $
$\Rightarrow { ax }^{ 2 }-2hxy+{ by }^{ 2 }=0 $

Let $Q(h,k)$ be the image of $O(0,0)$ w.r.t the line mirror $x+y+1=0$

$\displaystyle \frac{h-0}{1}=\frac{k-0}{1}=\frac{-2(1)}{2}$

$\displaystyle \Rightarrow h=k=-1$

$\Rightarrow h=-1, k=-1$
So, the image is at $(-1,-1)$

Reflecting the line y=x across 2x-y=1 involves reflecting two points on the line (e.g., (0,0) and (1,1)) across the mirror line and finding the equation of the line passing through the images.

Point A(1,2) reflected in y=x gives B(2,1) (coordinates swap). B(2,1) reflected in y=0 gives (2,-1) (y-coordinate negated). Therefore alpha=2, beta=-1. This is identical to question 463562, testing the same two-step reflection concept.

The point $P(2,1)$ if shifted by $3\sqrt2$ parallel to the line $x+y =1$, in the direction of increasing ordinate, to reach point $Q(\alpha, \beta)$.

Follow the steps: (4, 1) -> (1, 4) -> (3, 4) -> rotate by 45 deg -> reflection about x=0.

From given points, required line and line joining points are perpendicular to each other,

Reflect (1, 2) across the x-axis to get (1, -2). The line connecting (1, -2) and (5, 3) intersects the x-axis at the point of reflection A.

Use the reflection formula for a point (x1, y1) in line ax+by+c=0: (x-x1)/a = (y-y1)/b = -2(ax1+by1+c)/(a^2+b^2).

Here the mirror image of line $3x^2+4xy+5y^2=0$ in mirror $x=0$

Let $(h,k)$ be the point of reflection.
Line joining $(0,0)$ and $(h,k)$ is perpendicular to $4x-2y-5=0$
Product of slopes of these lines is $-1$
$\dfrac{k}h*2=-1=>k=-\dfrac{h}2$
Midpoint of $(0,0)$ and $(h,k)$ i.e, $(\dfrac{h}{2},\dfrac{k}{2})$ lies on $4x-2y-5=0$
Therefore $4\dfrac{h}2-2\dfrac{k}2-5=0=>2h-k-5=0$ but $k=-\dfrac{h}2$
$2h+\dfrac{h}2-5=0=>h=2 $ and $k=-1$
Point of reflection is $(2,-1)$.

The given equation of the circle is $x^2 + y^2 + 16x - 24y + 183 = 0$, which can be written as,

Under the transformation reflection about $y = x$ the new point is $(1,4)$
Under the transformation of shifting it $+2$ units along x-axis the new point is $(1+2,4) = (3,4)$
Under the transformation of rotation of $\dfrac{\pi}{4}$
$y= -X \sin  \theta +y\cos\ \theta$

$X=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{7}{\sqrt{2}}$

$y=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}$

Let the image or (reflection) of the origin with reference to the line $4x + 3y - 25 = 0$ is $(h, k)$.
$\therefore \dfrac {h - 0}{4} = \dfrac {k - 0}{3} = \dfrac {-2(0 + 0 - 25)}{16 + 9} = \dfrac {50}{25} = 2$
$\therefore \dfrac {h}{4} = 2\Rightarrow h = 8$
and $\dfrac {k}{3} = 2\Rightarrow k = 6$
$\therefore$ The required point is $(8, 6)$.

The incident ray reflects off x=-2. The reflected ray touches the circle x^2+y^2=4. Use the property that the incident ray, reflected ray, and normal at the point of incidence are related.

Image of any Point $(\alpha , \beta)$ with respect to line $ax+by +c = 0$ is given by equation,

Given Point $P(4,1)$

Here, $(x _1,y _1)$ is $(-8,12)$

The incident ray x+sqrt(3)y=sqrt(3) has slope -1/sqrt(3). It hits the x-axis at (sqrt(3), 0). The reflected ray will have slope 1/sqrt(3) and pass through (sqrt(3), 0). Equation: y-0 = 1/sqrt(3)(x-sqrt(3)) => sqrt(3)y = x-sqrt(3).

Reflecting (6, -1) across y=2: the x-coordinate stays 6. The distance from y=-1 to y=2 is 3. The new y-coordinate is 2+3=5.

The image of a point (x, y) in the y-axis is obtained by changing the sign of its x-coordinate while keeping the y-coordinate the same. For the point (2, -3), negating the x-coordinate gives (-2, -3).

Use the reflection formula for a point across a line to find the coordinates of P1 and P2 in terms of lambda and gamma, then solve for alpha.

Consider point P and Q
The line which is perpendicular to $y=-x$ and passes through $(a,b)$ will be
$\dfrac{y-b}{x-a}=1$
$y=x-(a-b)$
Now the point where it intersects $y=-x$ is
$-x=x-(a-b)$
$2x=(a-b)$
$x=\dfrac{a-b}{2}$ and hence $y=\dfrac{b-a}{2}$
This will be the mid point of PQ since Q is the image of point P on the line $y=-x$
Therefore $Q=(-b,-a)$
Similarly $R=(-a,-b)$
Hence The midpoint of PR will be $(0,0)$

Slope of line joining the image $Q(-2,6)$ and the point $P(4,2)$ is $\displaystyle -\frac{2}{3}$

So, the slope of mirror $L$ is $\displaystyle \frac{3}{2}$

Mid-point of $PQ$ is $(1,4)$

Since, the image and point are equidistant from mirror. So, this point $(1,4)$ lies on the mirror.

So, the equation of mirror is
$y-4=\displaystyle \frac{3}{2} (x-1)$

$\Rightarrow 3x-2y+5=0$

We have $y=|x-1|$