The line $3x-4y+7=0$ is rotated through an angle $\dfrac {\pi}{4}$ in clockwise direction about the point $\left (1,1\right)$. The equation of the line in its new position is

Let $\displaystyle A\equiv \left( 2,0 \right) $ and $\displaystyle B\equiv \left( 3,1 \right) $. The line $\displaystyle AB$ turns about $\displaystyle A$ through an angle $\displaystyle \frac { \pi  }{ 12 } $ in the clockwise sense, and the new position of $\displaystyle B$ is $\displaystyle B'$. Then $\displaystyle B'$ has the co-ordinates :-

Without changing the direction of coordinates axes,  origin is transferred to $(\alpha ,\ \beta)$ so that linear term in the equation $x^{2}+y^{2}+2x-4y+6=0$ are eliminated the point $(\alpha ,\ \beta)$ is   

The point $\mathrm{A}(2,1)$ is translated parallel to the line $x-y=3$ by a distance $4$ units. If the new position $A'$ is in third quadrant, then the coordinates of $A'$ are:

If the points $(5, 5), (7, 7)$ and $(a, 8)$ are collinear then the value of a is

${A}$ line has intercepts $ a$ and ${b}$ on the co ordinate axes. When the axes are rotated through an angle $\alpha$, keeping the origin fixed, the line makes equal intercepts on the coordinate axes, then $\tan\alpha=$ 

The angle of rotation of the axes so that the equation $\sqrt{3}\mathrm{x}-\mathrm{y}+5=0$ may be reduced to the form $\mathrm{Y}=\mathrm{k}$, where $\mathrm{k}$ is a constant is 

Find the equation of a line whose inclination is $\displaystyle 30^{\circ}$ and making an intercept of -3/5 on the y-axis

lf the equation $4\mathrm{x}^{2}+2\sqrt{3}\mathrm{x}\mathrm{y}+2\mathrm{y}^{2}-1=0$ becomes $5\mathrm{X}^{2}+\mathrm{Y}^{2}=1$, when the axes are rotated through an angle $\theta$, then $\theta$ is 

lf the distance between two given points is $2$ units and the points are transferred by shifting the origin to $(2, 2)$, then the distance between the points in their new position is.

If a point $\mathrm{P}(4,3)$ is shifted by a distances $\sqrt{2}$ units parallel to the line $\mathrm{y}=\mathrm{x}$, then the coordinates of $\mathrm{P}$ in its new position are

A line has intercepts a, b on the coordinate axes. If the axes are rotated about the origin through an angle $\displaystyle \alpha$ then the line has intercepts p,q on the new position of the axes respectively. Then 

If the origin is shifted to the point $(\displaystyle\frac{ab}{a-b}, 0)$ without rotation, then the equation $(a-b)(x^2 + y^2) - 2abx = 0$ becomes

The new equation of the curve $4(x-2y+1)^{2}+9(2x+y+2)^{2}=25$ if the lines $2x+y+2=0$ and $x-2y+1=0$ are taken as the new $x$ and $y$ axes respectively is

The coordinates axes are rotated about the origin $O$ in the counter clockwise direction through an angle of $\dfrac{\pi}{6}$. If $a$ and $b$ are intercepts made on the new axes by a straight line whose equation referred to old the axes is $x+y=1$, then the value of $\displaystyle \frac{1}{a^{2}}+\displaystyle \frac{1}{b^{2}}$ is equal to

The reflection of the plane $x+y+z-3=0$ in the plane $2x+3y+4z-6=0$

Reflection of the line $\dfrac{x-1}{-1}=\dfrac{y-2}{3}=\dfrac{z-4}{1}$ in the plane $x+y+z=7$ is:

The image of the line $x-y-1=0$ in the line $2x-3y+1=0$ is

The image of the point A$(1,2)$ by the line mirror y=x and the image of B by the line mirror $y=0$ is the point $\left(\alpha, \beta \right)$, then :

A ray of light travelling along the line $x+\sqrt{3}y=5$ is incident on the $x-axis$ and after refraction it enters the other side of the $x-axis$ by turning $\dfrac{\pi}{6}$ away from the $x-axis$. The equation of the line along which the refracted ray travels is

If $B$ is reflection of $A(a,5)$ about line $4x-3y=0$, then area of triangle $ABC$ is equal to

Locus of the image of the point (2, 3) in the line (2x - 3y + 4) + k(x - 2y + 3) = 0, k $\in $ R, is a 

The distance of the image of a point (or an object) from the line of symmetry (mirror) is  ----- as that of the point (object )from the line (mirror).

The image of the point (-5,4) under a reflection across the y-axis is (5,4).

Image of $\left (1,2\right)\ w.r.t\left (-2,-1\right)$ is

If the line $\left (2\cos \theta+ 3\sin \theta\right)$ $x+(\left (3\cos \theta- 5\sin \theta\right)$ $y-\left (5\cos \theta- 2\sin \theta\right)=0$ passes through a fixed point $P$ for all values $\theta$ and $Q$ be the image of the point $P$ with the respect to the line $4x+6y-23=0$, then the distance of $Q$ from the origin is:

The image of the point $A(1,2)$ by the line mirror  $y=x$ is the 
Point B and the image of B by the line mirror $y=0$ is the point $(a,\beta )then:$ 

A ray of light along $x + \sqrt {3y}  = \sqrt 3 $ gets reflected upon reaching $x - axis$ , then equation of the reflected ray is 

The line segment joining $A\left( {3,\,\,0} \right),\,\,B\left( {5,\,\,2} \right)$ is rotated about a point A in anticlockwise sense through an angle $\displaystyle{\pi  \over 4}$ and B move to C. If a point D be the reflection of C in y-axis, then D=

The reflection of the point $(2, -1, 3)$ in the plane $3x-2y-z=9$ is?

Find the image of the point $\displaystyle \left ( -2, -7 \right )$ under the transformation
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right ).$

If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to the line $L =$ $0$, then $L =$

The image of the pair of lines represented by$\displaystyle :ax^{2}+2hxy+by^{2}= 0 $ by the line $ y= 0 $ mirror is:

The coordinates of the image of the origin $O$ with respect to the line $x+y+1=0$ are

The equation of the line AB is y = x. if And B lie on the same side of the line mirror 2x - y = 1, then the equation of the image of AB is   _____________.

The image of the point $A(1, 2)$ by the line mirror $y=x$ is the point $B$ and the image of $B$ by the line mirror $y=0$ is the point $(\alpha, \beta)$, then?

$P(2,1)$ is image of the point $Q(4,3)$ about the line

The point $P(2, 1)$ is shifted by $\displaystyle 3\sqrt{2}$ parallel to the line $\displaystyle x+y=1,$ in the direction of increasing ordinate, to reach $Q$.The image of $Q$ by the line $\displaystyle x+y=1$ is

The image of the line $\displaystyle  \frac { x - 1 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 4 } { - 5 } $ in the plane $2 x - y + z + 3 = 0 $ is the line

The point (4, 1) undergoes the following transformation successively.
(i)reflection about the line y=x
(ii)translation through a distance 2 units along the positive direction of x-axes.
(iii)rotation through an angle ${ \pi  }/{ 4 }$ about the origin in the anticlockwise direction.
(iv) reflection about x=0
The final position of the given point is

Let  $ABC$ be triangle. Let $A$ be the point $(1,2),y=x$be the perpendicular bisector of $AB$ and $x-2y+1=0$ be the angle bisector of $\angle C$. If equation of $BC$ is given by $ax+by-5=0$, then the value of $a+b$ is 

If the image of the point $ \displaystyle \left ( 4,-6 \right ) $ by a line is the point $(2,2)$, then the equation of the mirror is

A ray light comming from the point $(1,2)$ is reflected at a point $A$ on the $x-$axis and then passes through the point $(5,3)$. The co-ordinates of the point $A$ is

The point $A(4, 1)$ undergoes following transformations successively:
(i) reflection about line $y=x$
(ii) translation through a distance of $3$ units in the positive direction of x-axis.
(iii) rotation through an angle $105^o$ in anti-clockwise direction about origin O.
Then the final position of point A is?

The reflection of the point $(4, -13)$ in the line $5x+y+6=0$ is

The image of the pair of lines represented by $\displaystyle 3x^{2}+4xy+5y^{2}=0 $ in the line mirror $x = 0$ is

The point A(4, 1) undergoes following transformations successively
(i) reflection about line y = x
(ii) translation through a distance of 2 units in the positive direction of x axis
(iii) rotation through an angle $\displaystyle \pi/4 $ in anti clockwise direction about origin O
Then the final position of point A is

The co-ordinates of the point of reflection of the origin $(0, 0)$ in the line $4x -2y - 5 = 0$ is

The equation of the image of the circle $\displaystyle x^{2}+y^{2}+16x-24y+183=0 $ along the line mirror $4x + 7y + 13 = 0$ is:

The image of the pair of lines represented by $\displaystyle  3x^{2}+4xy+5y^{2}=0 $ in the line mirror x = 0 is

Let $0<\alpha< \dfrac{\pi}{4}$ be a fixed angle. If $\mathrm{P}=(\cos\theta,\sin\theta)$ and $\mathrm{Q}=(\cos(\alpha-\theta),\sin(\alpha-\theta))$ then $\mathrm{Q}$ is obtained from $\mathrm{P}$ by :

The point $(4, 1)$ undergoes the following three transformations successively
i) Reflection about the line $\mathrm{y}=\mathrm{x}$
ii) Transformation through a distance of $2$ units along the $+\mathrm{v}\mathrm{e}$ direction of the x-axis
iii) Rotation through an angle $\displaystyle \frac{\pi}{4}$ about the origin in the anticlockwise direction. The final position of the point is given by the co-ordinates 

The image of the origin with reference to the line $4x + 3y - 25 = 0$, is

A light ray gets reflected from the $ x= -2 $ .If the reflected ray touches the circle $ x^{2}+y^{2}=4 $ and point of incident is $(-2,-4)$,then equation of incident ray is

The image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ is

 The point $(4, 1)$ undergoes the following three transformations successively
(a) Reflection about the line $y = x$

(b) Transformation through a distance $2$ units along the positive direction of the x-axis.

(c) Rotation through an angle $p/4$ about the origin in the anti clockwise direction.

The final position of the point is given by the co-ordinates

Image of the point $\left( -8,12 \right) $ with respect to the line mirror $4x+7y+13=0$ is 

Equation of line equidistant from lines $2x + 3y = 5$ and $4x + 6y = 11$ is

A ray of light along $x+\sqrt{3}y=\sqrt{3}$ get reflected upon reaching x-axis, the equation of the reflected ray is?

What is the reflection of the point $(6,-1)$ in the line $y=2$?

The image of (2, -3) in the y - axis is

If ${ P } _{ 1 }\left( \dfrac { 1 }{ 5 } ,\alpha  \right)$ and ${P } _{ 2 }\left( \beta ,\dfrac { 18 }{ 5 }  \right)$ be the images of point $P\left( 1,\gamma  \right)$ about lines ${ L } _{ 1 }:2x-y=\lambda$ and ${ L } _{ 2 }:2y+x=4$ respectively, then the value of $\alpha$is-

The image of $P(a, b)$ in the line $y= -x$ is $Q$ and the image of $Q$ in the line $y=x$ is $R$. Then the midpoint of $PR$ is

If $\displaystyle \left ( -2, 6 \right )$ is the image of the point $\displaystyle \left ( 4,2 \right )$ with respect to the line $\displaystyle L=0$, then $\displaystyle L=$

The equation of image of pair of lines $y=|x-1|$ with respect to y-axis is