Tag: maths

Questions Related to maths

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:

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

  1. $(2+2\sqrt{2},2+2\sqrt{2})$

  2. $(-2+\sqrt{2},-1-2\sqrt{2})$

  3. $(2-2\sqrt{2},1-2\sqrt{2})$

  4. $(-2-\sqrt{2},-1-2\sqrt{2})$

Show answer
Correct Option: C
Explanation:

$ 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 $

$x =2 \pm 4\times \dfrac{1}{\sqrt{2}} ;\ y=1 \pm 4\times \dfrac{1}{\sqrt{2}}$.....(consider -ive sign for third quadrant)
$x=2-2\sqrt{2};\ y=1-2\sqrt{2}$ since they are in third quadrant.

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

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

  1. $6$

  2. $3$

  3. $8$

  4. $7$

Show answer
Correct Option: C
Explanation:

When three points are collinear, Slope of line joining any two points is same as the slope of line joining any other two points


Slope of line joining two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $\dfrac { {y} _{2} - {y} _{1}}{ {x} _{2} - {x} _{1}} $

So, Slope of line joining $ (5,5) ;  (7,7) $ is $ \dfrac {7-5}{7-5} = \dfrac {2}{2} = 1 $

And Slope of line joining $ (a,8) ;  (7,7) $ is $ \dfrac {7-a}{7-8} = a - 7 $

As they are collinear $ a - 7 = 1 => a = 8 $



${A}$ 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 $\tan\alpha=$ 

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

  1. $\displaystyle \frac{{a}+b}{{a}-b}$

  2. $\displaystyle \frac{{a}-b}{{a}+b}$

  3. $\dfrac{b}{a}$

  4. $\displaystyle \frac{{a}}{b}$

Show answer
Correct Option: B
Explanation:
Let the equation of line be $ \displaystyle \frac{x}{a}+\frac{y}{b}=1.$

When axes are rotated through an angle $ \alpha$, the new coordinates $XY$ are related to old coordinates $xy$ as follows:

$x=X \cos\alpha -Y\sin \alpha $

$y=X \sin\alpha +Y\ \cos \alpha $

Substituting these values in the equation of line, we get
$ \displaystyle \frac { X\cos { \alpha  } -Y\sin { \alpha  }  }{ a } +\frac { X\sin { \alpha  } +Y\cos { \alpha  }  }{ b } =1\\ \displaystyle \Rightarrow X\left( \frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b }  \right) +Y\left( \frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }  \right) =1$

As it makes equal intercepts in the new coordinate system, we get

$ \displaystyle \Rightarrow \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$

$\Rightarrow \cos \alpha \left ( b-a \right )=-\sin \alpha \left ( a+b \right )$

$ \displaystyle \Rightarrow \tan  \alpha =\dfrac{a-b}{a+b}$

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 

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

  1. $\dfrac{\pi}{6}$

  2. $\dfrac{\pi}{4}$

  3. $\dfrac{\pi}{3}$

  4. $\dfrac{\pi}{12}$

Show answer
Correct Option: C
Explanation:
Let axis be rotated through an angle $\theta $ then 
$ x= x^{1} \cos \theta - y^{1} \sin\theta $
$ y = x^{1} \sin \theta + y^{1} \cos\theta $
$ \sqrt{3}\times x -y +5 = 0$
$ \sqrt{3} (x^{1} \cos \theta - y^{1} \sin\theta) - (x^{1} \sin \theta + y^{1} \cos\theta) +5 = 0$
$x^{1} (\sqrt{3} \cos \theta - \sin\theta ) = 0$
$ \Rightarrow \tan\theta  = \sqrt{3}$
$ \theta = 60^{\circ} = \dfrac{\pi }{3}$

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

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

  1. $ y = \dfrac {5}{\sqrt{6}}x  +\dfrac {2}{5} $

  2. $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

  3. $ y = \dfrac {3}{\sqrt{7}}x  -\dfrac {1}{3} $

  4. $\displaystyle \dfrac{-5}{3}x$

Show answer
Correct Option: B
Explanation:

The equation of any straight line can be written as $ y = mx + c $, where $m$ is its slope and $c$ is its y - intercept.

As inclination is $ 30^o $, slope of the line $ =  tan (30 ^o) = \dfrac {1}{\sqrt{3}} $

So equation of line is $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

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 

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

  1. $15^{\mathrm{o}}$

  2. $30^{\mathrm{o}}$

  3. $45^{0}$

  4. $60^{\mathrm{o}}$

Show answer
Correct Option: B
Explanation:
By  rotation  of  axes  through  $\theta $,  co-ordinates  become
$x = x^{1}  \cos\theta  - y^{1}  \sin\theta $
$y = x^{1}  \sin\theta  + y^{1}  \cos\theta $
$4x^{2} + 2\sqrt{3}xy + 2y^{2} - 1=0$
$\Rightarrow 4(x^{1}\cos\theta - y^{1} \sin\theta )^{2} + 2\sqrt{3} (x^{1} \cos\theta  -y^{1} \sin\theta )  (x^{1} \sin\theta + y^{1} \cos\theta )  +2 (x^{1}\sin\theta +y^{1} \cos\theta )^{2} -1 =0$
coeff  of $xy=0$
$\Rightarrow 4(-\sin2\theta )+2\sqrt{3}   \cos20  +  2 \sin2\theta  = 0$
$2\sqrt{3}   \cos2\theta  = 2\sin2\theta $
$\tan2\theta = \sqrt{3}$
$2\theta = \dfrac{\pi }{3}$
$\theta  = \dfrac{\pi }{6} = 30^{\circ}$
$\therefore $ angle  to  be  rotated $= 30^{\circ}$

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.

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

  1. $2$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: A
Explanation:

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.

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

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

  1. $(5,4)$

  2. $(5+\sqrt{2},4+\sqrt{2})$

  3. $(5-\sqrt{2},4-\sqrt{2})$

  4. $(4,5)$

Show answer
Correct Option: A
Explanation:

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

for the line is tan $\theta =1$

$\therefore$ by parametric equations, we have

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

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

$x=5$  ;  $y=4$

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 

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

  1. $\displaystyle \frac{1}{p^{2}}+\frac{1}{q^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

  2. $\displaystyle \frac{1}{p^{2}}-\frac{1}{q^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}$

  3. $\displaystyle \frac{1}{p^{2}}+\frac{1}{a^{2}}=\frac{1}{q^{2}}+\frac{1}{b^{2}}$

  4. None of these

Show answer
Correct Option: A
Explanation:

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 } } $

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

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

  1. $(a-b) (X^2+Y^2) - (a+b)XY + abX = a^2$

  2. $(a+b) (X^2 + Y^2) = 2ab$

  3. $(X^2 + Y^2) = (a^2 + b^2)$

  4. $(a-b)^2 (X^2 + Y^2) = a^2 b^2$

Show answer
Correct Option: D
Explanation:

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
$(a-b) \displaystyle \left [ \left (X + \frac{ab}{a-b} \right )^2 + y^2 \right ] - 2ab \left [ X+ \frac{ab}{a-b} \right ] = 0$
$\Rightarrow (a-b) \displaystyle \left [ X^2 + \frac{a^2 b^2}{(a-b)^2} + Y^2 + \frac{2abX}{a-b} \right ] - 2abX - \frac{2a^2 b^2}{a-b} = 0$
$\Rightarrow (a-b) (X^2 + Y^2) = \displaystyle \frac{a^2 b^2}{a-b}$
$\Rightarrow (a-b)^2 (X^2+Y^2) = a^2 b^2$