Mathematics

Straight Lines and Angles

162 Questions

Straight lines and angles are core components of coordinate geometry. This topic evaluates angle measures between intersecting lines, direction ratios, and perpendicular distances. Mastery of these mathematical concepts is necessary for high scores in quantitative exams.

Angle between linesDirection ratiosAngle bisectorsPerpendicular distanceSlope differences

Straight Lines and Angles Questions

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The pair of lines $6{ x }^{ 2 }+7xy+\lambda { y }^{ 2 }=0\left( \lambda \neq -6 \right) $ forms a right angled triangle with $x+3y+4=0$ then $\lambda=$

  1. $3$
  2. $-3$
  3. $1/3$
  4. $-1/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given line is $L: x+3y+4=0$


$\implies  y=-\dfrac{1}{3}(x+4)$

Slope of this line is $m=\dfrac{-1}{3}$

Now, $6x^2+7xy+\lambda y^2=0$ $(\lambda\neq 6)$

$x^2+\dfrac{7}{6}xy+\dfrac{\lambda}{6}y^2=0$

$\implies (x+ay)(x+by)=0$

$\implies x+ay=0$ and $x+by=0$ are the two equations with 

$a+b=\dfrac{7}{6}$    and $ab=\dfrac{\lambda}{6}$

Slope of these lines are $m _1=\dfrac{-1}{a}$ and $m _2=\dfrac{-1}{b}$

Now, $m _1m _2=\dfrac{1}{ab}=\dfrac{\lambda}{6}\neq -1$  since $\lambda\neq -6$

Hence the lines $x+ay=0$ and $x+by=0$ are not prependicular.

From these two only one is normal to $L$.

Let $x+ay$ is normal to $L$.

$\implies m _1m=-1$

$\implies \dfrac{1}{3a}=-1$

$\implies a=\dfrac{-1}{3}$

Now, $a+b=\dfrac{7}{6}\implies b=\dfrac{7}{6}-\dfrac{-1}{3}$

$\implies b=\dfrac{3}{2}$

Now, $ab=\dfrac{\lambda}{6}$

$\implies \lambda=6ab=6\times \dfrac{-1}{3}\times \dfrac{3}{2}$

$\implies \lambda=-3$

Answer-(B)

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Assertion(A): The angle between the asymptotes of $3x^{2}-y^{2}=3$ is $120^{\circ}$
Reason(R): The angle between the asymptotes of $x^{2}-y^{2}=a^{2}$ is $90^{\circ}$

  1. Both A and R are true and R is the correct

    explanation of A.

  2. Both A and R are true but R is not correct

    explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Asymptotes of a hyperbola is given by $y=\pm \frac { x }{ a } $


So for hyperbola $3{ x }^{ 2 }-{ y }^{ 2 }=3$
The asymptotes make an angle of ${ 60 }^{ o }$ and  ${ 120 }^{ o }$ with x-axis which means they make an angle of ${ 60 }^{ o } $ among themself.

Now for hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={a}^{2}$

The asymptotes makes and angle of ${45}^{o}$ and  ${135}^{0}$ which means ${90}^{o}$ among themself.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The following lines are $\hat { r } =\left( \hat { i } +\hat { j }  \right) +\lambda \left( \hat { i } +2\hat { j } -\hat { k }  \right) +\mu \left( -\hat { i } +\hat { j } -\hat { 2k }  \right) $

  1. collinear

  2. skew-lines

  3. co-planar lines

  4. parallel lines
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Condition for three lines $\vec { { r } _{ 1 } } $ , $\vec { { r } _{ 2 } } $ , and $\vec { { r } _{ 3 } } $ to be collinear is:

$\vec { { r } _{ 1 } } +\lambda \vec { { r } _{ 2 } } +\vec { { \mu r } _{ 3 } } =0$
where $\vec { { r } _{ 1 } } =\left( \vec { i } +\vec { j }  \right) $
$\vec { { r } _{ 2 } } =\left( \vec { i } +2\vec { j } -\vec { k }  \right) $
$\vec { { r } _{ 3 } } =\left( -\vec { i } +\vec { j } -2\vec { k }  \right) $
and $\lambda $ and $\mu $ are scalars
Hence, the answer is collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the lines $x=1+a,y=-3-\lambda a,z=1+\lambda a$ and $x=\cfrac { b }{ 2 } ,y=1+b,z=2-b$ are coplanar, then $\lambda$ is equal to

  1. $-3$
  2. $2$
  3. $1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The given lines are $\cfrac { x-1 }{ 1 } =\cfrac { y+3 }{ -\lambda  } =\cfrac { z-1 }{ \lambda  } =\left( a \right) $ and $\cfrac { x-0 }{ 1/2 } =\cfrac { y-1 }{ 1 } =\cfrac { z-2 }{ -1 } =(b)$
$\therefore$ coplanarity, we must have
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda  & \lambda  \\ 1/2 & 1 & -1 \end{vmatrix}=0$
$\Rightarrow -1\left( \lambda -\lambda  \right) -4\left( -1-\cfrac { \lambda  }{ 2 }  \right) +\left( 1+\cfrac { \lambda  }{ 2 }  \right) =0$
$4+2\lambda +1+\cfrac { \lambda  }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda  }{ 2 } =0\quad \therefore \lambda =-2$

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

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

  1. $7y+x-6=0$
  2. $7y-x-6=0$
  3. $x+7y=8$
  4. $7y-x+6=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given slope $=\dfrac{3}{4} < 1$
$< 45^{o}$
$\therefore $ after rational clock slope because negative
$\dfrac{m _1-m _{2}}{1+m _{1},m _{2}}=\tan 45$
$m _{1}=\dfrac{3}{4} \,\,\,m _{2}=$ new slope
$\left| \dfrac{\dfrac{3}{4}-m}{1+\dfrac{3}{4} m} \right|=1$
$\dfrac{3}{4}-m= \pm \left( 1+\dfrac{3}{4}m \right)$
$\dfrac{3}{4}-m=1+ \dfrac{3}{4} m$
$\dfrac{7}{4}m=\dfrac{-1}{4}$
$m=\dfrac{-1}{7}$ appeared 
$\therefore$ satisfy slope and pt in option to save time.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

${A}$ 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 $\tan\alpha=$ 

  1. $\displaystyle \frac{{a}+b}{{a}-b}$
  2. $\displaystyle \frac{{a}-b}{{a}+b}$
  3. $\dfrac{b}{a}$
  4. $\displaystyle \frac{{a}}{b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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 

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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 

  1. $15^{\mathrm{o}}$
  2. $30^{\mathrm{o}}$
  3. $45^{0}$
  4. $60^{\mathrm{o}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= 30^{\circ}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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 

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } } $

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

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

  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. $(a-b)^2 (X^2 + Y^2) = a^2 b^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$(a-b) \displaystyle \left [ \left (X + \frac{ab}{a-b} \right )^2 + y^2 \right ] - 2ab \left [ X+ \frac{ab}{a-b} \right ] = 0$
$\Rightarrow (a-b) \displaystyle \left [ X^2 + \frac{a^2 b^2}{(a-b)^2} + Y^2 + \frac{2abX}{a-b} \right ] - 2abX - \frac{2a^2 b^2}{a-b} = 0$
$\Rightarrow (a-b) (X^2 + Y^2) = \displaystyle \frac{a^2 b^2}{a-b}$
$\Rightarrow (a-b)^2 (X^2+Y^2) = a^2 b^2$

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

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

  1. $1$
  2. $2$
  3. $4$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

$\displaystyle\frac{X}{a}+\frac{Y}{b}=1$ .....(ii)
$\displaystyle\Rightarrow \frac { 1 }{ a } =\frac { (\sqrt { 3 } +1) }{ 2 } $
$\Rightarrow \displaystyle \frac { 1 }{ b } =\frac { (1-\sqrt { 3 } ) }{ 2 } $
So, $\displaystyle \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } =2$

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

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

  1. $\left ( \displaystyle -\frac{1}{2},\displaystyle -\frac{1}{2} \right )$
  2. $(-2,-2)$
  3. $(1,1)$
  4. $(-1,-1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 $(-1,-1)$

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

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 :

  1. clockwise rotation around the origin through an angle $\alpha$
  2. anticlockwise rotation around the origin through an angle $\alpha$
  3. reflection in the line through origin with slope $\tan\alpha$
  4. reflection in the line through origin with slope $\displaystyle \tan\frac{\alpha}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$P =\left(\cos \theta, \sin \theta\right)$ ; $Q=\left(\cos \left(\alpha -\theta \right),\sin \left(\alpha -\theta \right)\right)$

Angle between $P$ and $Q$ is $\tan^{-1}$  $\left(\dfrac{\sin\left(\alpha -\theta \right)-\sin\alpha }{\cos\left(\alpha -\theta \right)-\cos\alpha }  \right)$

$=\tan^{-1}\left( \tan \alpha  \right)=\alpha $

$\therefore$ mid point of $P$ and $Q$ is $ \left(\dfrac{\cos\theta + \cos(\alpha -\theta)}{2}, \dfrac{\sin\theta +\sin( \alpha -\theta)}{2}\right)$

$= \left(  \cos\dfrac{\alpha }{2} \cos\left(\dfrac{\theta-\alpha }{2}\right), \sin\dfrac{\alpha }{2}\cos\left(\dfrac{\theta-\alpha }{2}\right) \right)$

$= \cos\left(\dfrac {\theta -\alpha }{2}\right)\left( \cos\dfrac{\alpha }{2},\sin\dfrac{\alpha }{2} \right)$

Which is a point on line with slope $y=\tan\dfrac{\alpha }{2}$

$\therefore$ $Q$ is obtained by reflection of origin with slope $\tan$ $\dfrac{\alpha }{2}$.
Hence, option 'D' is correct.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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

  1. $(-8, 6)$
  2. $(8, 6)$
  3. $(-3, 4)$
  4. $(8, -6)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $(8, 6)$.