Mathematics

Straight Lines and Angles

122 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 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

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   

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

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).

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 $(8, 6)$.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum of the intercepts cut off by the axes on the lines  $ x+y=a,x+y=ar,x+y=ar^{2}\ldots\ldots\ldots$ where $a\neq 0$ and $r=\displaystyle \dfrac{1}{2}$  is 

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

$x+y=a$

intercept cut off by the axes $=\sqrt{a^2+a^2}=\sqrt{2}a$

$\therefore$ sum of all intercepts cut off by the axes on the lines,

$x+y=a, x+y=ar,.... x+y=a^{r^n-1}$

$\sqrt{2}a, \sqrt{2}ar,.... \sqrt{2}ar^{n-1}$

$Sum=\sqrt{2}a+\sqrt{2}ar+....+\sqrt{2}a^{r^n-1}....$as

$=\sqrt{2}(\dfrac{a}{1-r})$

$=\dfrac{\sqrt{2}a}{1-r}$

Given  $\Rightarrow r=\dfrac{1}{2}$

$\therefore Sum=2\sqrt{2}a$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the lines $\frac{x - 0}{1} =\frac{y+1}{2}=\frac{z-1}{-1}$ and $\frac{x+1}{k}=\frac{y-3}{-2}=\frac{z-2}{1}$ are at right angles, then the value of k is

  1. $5$
  2. $0$
  3. $3$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $({ l } _{ 1 }{ m } _{ 1 }{ n } _{ 1 })$ and  $({ l } _{ 2 }{ m } _{ 2 }{ n } _{ 2 })$ are directions of two $\bot$ lines then,

${ l } _{ 1 }{ l } _{ 2 }+{ m } _{ 1 }{ m } _{ 2 }+n _{ 1 }{ n } _{ 2 }=0\ k-4-1=0\ k=5$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Perpendicular distance from the origin to the line joining the points $(a\cos{\theta},a\sin{\theta})(a\cos{\theta},a\sin{\theta})$ is

  1. $2a\cos{(\theta-\phi)}$
  2. $a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $
  3. $4a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $
  4. $a\cos { \left( \cfrac { \theta +\phi }{ 2 } \right) } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The difference of the slopes of the lines $x ^ { 2 } \left( \sec ^ { 2 } \theta - \sin ^ { 2 } \theta \right) - ( 2 \tan \theta ) x y + y ^ { 2 } \sin ^ { 2 } \theta = 0$

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

According to the question..........

$\begin{array}{l} Let,\, { m _{ 1\,  } }&amp; \, { m _{ 2 } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ and,\,  \ product\, of\, slope:{ m _{ 1 } }.\, { m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ Here, \ a{ x^{ 2 } }+2hxy+b{ y^{ 2 } }=0.........(general\, equ\, of\, straight\, line.) \ cofficient\, of: \ a={ \sec ^{ 2 }  }\theta -{ \sin ^{ 2 }  }\theta  \ h=-\tan  \theta  \ b={ \sin ^{ 2 }  }\theta  \ Now,\, value\, put\, { { into } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ \Rightarrow { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2(-tan\theta ) } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { { 2\sin  \theta \times 2 } }{ { 2{ { \sin   }^{ 2 } }\theta \, .\, \cos  \theta  } } =\dfrac { 4 }{ { 2sin\theta \cos  \theta  } } =\dfrac { 4 }{ { \sin  2\theta  } }  \ and, \ product\, of\, slope:{ m _{ 1 } }+{ m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ \Rightarrow { m _{ 1 } }.\, { m _{ 2 } }=\dfrac { { { { \sec   }^{ 2 } }\theta -{ { \sin   }^{ 2 } }\theta  } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { 1 }{ { { { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, \, \, \, \, \, \, \left[ { divide\, by\, 4 } \right.  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { 4 }{ { 4{ { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, =\dfrac { 4 }{ { { { (\sin  2\theta ) }^{ 2 } }\,  } } -1\,  \ \, \, \, Now,find\, difference: \ \, \, \, \, \, \, \, \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }={ ({ m _{ 1 } }+{ m _{ 2 } })^{ 2 } }-4{ m _{ 1 } }.\, { m _{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ={ \left( { \dfrac { 4 }{ { \sin  2\theta  } }  } \right) ^{ 2 } }-4\left( { \dfrac { 4 }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -1\,  } \right)  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -\, \dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } +4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }\, \, \, =4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, ({ m _{ 1 } }-{ m _{ 2 } })=+\sqrt { 4 } =2 \ \, \, \, \therefore \, \, \, the\, \, differece\, of\, slope\, \, is\, 2. \ So,\, that\, the\, correct\, option\, is\, B.\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  \end{array}$