Tag: position and movement

Questions Related to position and movement

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 

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} { m _{ AB } }=-1 \ y-2=-1\left( { x-1 } \right)  \ y-2=-x+1 \ x+y=3 \ \underline { x-y=0 }  \ 2x=3 \ \therefore x=\frac { 3 }{ 2 } \, \, \, \, \, ,y=\frac { 3 }{ 2 }  \ \frac { { 1+h } }{ 2 } =\frac { 3 }{ 2 } \, \, \,  \ h=2 \ \frac { { 2+k } }{ 2 } =\frac { 3 }{ 2 }  \ k=1\, \, \, \, \, \, \, \, \, \, \, B\left( { 2,1 } \right)  \ As\, \, image\, \, through\, \, x-2y+1=0 \ { m _{ AD } }=-2 \ y-2=-2\left( { x-1 } \right)  \ y-2=-2x+2 \ 2x+y=4\times 2\, \, \, \, \, \, \, x-2y+1=0 \ 4x+2y=8 \ \underline { x-2y=-1 }  \ 5x=7 \ x=\frac { 7 }{ 5 } \, \, \, \, \, \, y=\frac { { x+1 } }{ 2 } =\frac { { 12 } }{ { 5\times 2 } } =\frac { 6 }{ 5 }  \ \frac { { m+1 } }{ 2 } =\frac { 7 }{ 5 } \, \, \, \, \, \, \, \, \frac { { n+1 } }{ 2 } =\frac { 6 }{ 5 }  \ m+1=\frac { { 14 } }{ 5 } \, \, \, \, \, \, \, n+2=\frac { { 12 } }{ 5 }  \ m=\frac { 9 }{ 5 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n=\frac { { 12 } }{ 5 } -2=\frac { 2 }{ 5 }  \ E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ BC=BE\, \, \, \, \, \, B\left( { 2,1 } \right) \, \, \, E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ y-1=\left( { \frac { { 1-\frac { 2 }{ 5 }  } }{ { 2-\frac { 9 }{ 5 }  } }  } \right) \left( { x-2 } \right)  \ 3x-y=5 \ a=3\, \, \, \, b=-1 \ a+b=2 \end{array}$

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $ \displaystyle 4x+3y-5= 0 $

  2. $ \displaystyle x-4y= 11 $

  3. $ \displaystyle x+4y-5= 0 $

  4. $ \displaystyle -x+y+11= 0 $

Show answer
Correct Option: B
Explanation:

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


$\displaystyle \therefore$ slope of required line is $\displaystyle m=-\frac { 2-4 }{ 2-(-6) } =\frac { 1 }{ 4 } $

The midpoint of the given points will be on our line.

$\Rightarrow$$X=\left ( \dfrac{x _1+x _2}{2}\right ) and \ Y=\left ( \dfrac{y _1+y _2}{2}\right )$

$\Rightarrow$$X=\left ( \dfrac{4+2}{2}\right ) =3 \ and \ Y=\left ( \dfrac{-6+2}{2}\right )=-2$

$\displaystyle \therefore$ point is $\displaystyle (3,-2)\equiv(x _1,y _1)$.

$\therefore$ Equation of line is given by, 

$\Rightarrow$$y-y _1=m(x-x _1)$

Then equation of required line is $\displaystyle \frac { y+2 }{ x-3 } =\frac { 1 }{ 4 } \Longrightarrow x-4y=11$

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\left(\dfrac {13}{5}, 0\right)$

  2. $\left(\dfrac {5}{13}, 0\right)$

  3. $(-7, 0)$

  4. $None\ of\ these$

Show answer
Correct Option: A

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?

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\left(\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$

  2. $(-2, 7\sqrt{2})$

  3. $\left(-\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$

  4. $(-2\sqrt{6}, 2\sqrt{2})$

Show answer
Correct Option: A

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $(-1, -14)$

  2. $(3,4)$

  3. $(1,2)$

  4. $(-4, 13)$

Show answer
Correct Option: A

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle 3x^{2}-4xy+5y^{2}=0 $

  2. $\displaystyle 3x^{2}-4xy-5y^{2}=0 $

  3. $\displaystyle 5y^{2}-4xy-3x^{2}=0 $

  4. none of these

Show answer
Correct Option: A
Explanation:

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

i.e about $Y-axis $
So the x-coordinates About $Y-axis $ becomes negative and the y-coordiantes remains same Hence putting $x=-x$ in given eq 
$3(-x)^2-4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle \left ( \frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}} \right ) $

  2. $\displaystyle \left ( -2,7\sqrt{2} \right )$

  3. $\displaystyle \left ( -\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}} \right )$

  4. none of these

Show answer
Correct Option: C
Explanation:
Given Point $A(4,1)$
$(i) $ reflection about $y=x$
Hence putting coordinates of point A in given eq $y=4,x=1$
The point becomes $A(1,4)$

$(ii)$ translation through distance of 2 units in positive X axis 
$A(1+2,4)\Rightarrow A(3,4)$

$(iii)$ rotation of point through and angle $\dfrac{\pi}{4}$ in anticlockwise about origin O
After rotation Point will be $A(r\cos(\alpha+\dfrac{pi}{4}),r\sin(\alpha+\dfrac{pi}{4}))$
Converting into polar form 
$r=\sqrt{3^2+4^2}=5$
$\tan\alpha=\dfrac{4}{3}$
Hence $\cos\alpha=\dfrac{3}{5}$
$\sin\alpha=\dfrac{4}{5}$
$\cos(\alpha+\dfrac{\pi}{4})=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{3}{5}\dfrac{1}{\sqrt{2}}-\dfrac{4}{5}\dfrac{1}{\sqrt{2}}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{-1}{5\sqrt{2}}$

$\sin(\alpha+\dfrac{\pi}{4})=\sin\alpha\cos\dfrac{\pi}{4}+\cos\alpha\sin\dfrac{\pi}{4}$
$\sin(\alpha+\dfrac{\pi}{4})=\dfrac{4}{5}\dfrac{1}{\sqrt{2}}+\dfrac{3}{5}\dfrac{1}{\sqrt{2}}$
$\sin(\alpha+\dfrac{\pi}{4})=\dfrac{7}{5\sqrt{2}}$
$A\left(5\times \dfrac{-1}{5\sqrt{2}},5\times \dfrac{7}{5\sqrt{2}}\right)$
$A\left(-\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $(-1, 2)$

  2. $(2, -1)$

  3. $\displaystyle \left (\frac {4}{5}, -\frac {2}{5}\right )$

  4. $(2, 5)$

Show answer
Correct Option: B
Explanation:

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 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:

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle x^{2}+y^{2}+32x-4y+235=0 $

  2. $\displaystyle x^{2}+y^{2}+32x+4y-235=0 $

  3. $\displaystyle x^{2}+y^{2}+32x-4y-235=0 $

  4. $\displaystyle x^{2}+y^{2}+32x+4y+235=0 $

Show answer
Correct Option: D
Explanation:

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


$\Rightarrow (x+8)^2 + (y-12)^2 = (5)^2$

Hence we can see the center of the circle , let's say $O(- 8, 12)$ and radius of the circle is $r = 5$

If we mirror the image of the given circle by the line $4x +7y +13 =0$, the radius of the circle won't change. But the position of center will get change. Let's assume the new center will be $O'( \alpha, \beta)$

By using below equation to find $O'(\alpha, \beta)$, 

$\Rightarrow \dfrac{\alpha -(-8 )}{4} = \dfrac{\beta -12}{7} = \dfrac {-2 (4(-8) + 7(12) +13)}{4^2 +7^2} $

$\Rightarrow \dfrac{\alpha -(-8 )}{4} = \dfrac{\beta -12}{7} = -2$

$\Rightarrow \alpha = -16, \beta = -2$

Hence new center $O'$ is $O'(-16,-2)$

The equation of the image of the circle through the mirror will be,

$\Rightarrow (x+16)^2 + (y +2)^2 = (5)^2$

$\Rightarrow x^2 +y^2 + 32x +4y + 235$

So correct option is $D$

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

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle 3x^{2}-4xy+5y^{2}=0 $

  2. $\displaystyle 3x^{2}-4xy-5y^{2}=0 $

  3. $\displaystyle 5y^{2}-4xy-3x^{2}=0 $

  4. none of these

Show answer
Correct Option: A
Explanation:
Given pair
$3x^2+4xy+5y^2=0$
$x=0$ is the $Y-axis $ hence the $x$-coordinates will become $-x$ and $y$-coordinates remains same 
Hence 
$3(-x)^2+4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$