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 two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the equation of the pair of straight lines passing through the point $(1, 1)$, one making an angle $\theta$ with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis, is $x^2 - (a + 2)xy + y^2 + a(x + y -1) =0,   a  \neq 2$, then the value of sin 2$\theta$ is

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

The lines will be
$y-1=\tan A(x-1)$
and $y-1=\cot A(x-1)$
Therefore their joint equation will be
$(y-1-\cot A(x-1))(y-1-\tan A(x-1))=0$
$(y-1)^{2}-(\cot A+ \tan A)(x-1)(y-1)+(x-1)^{2}=0$
$y^2-2y+1-(\cot A+\tan A)(xy-x-y+1)+(x^2-2x+1)=0$
$x^2+y^2-(\cot A+\tan A)(xy)+((\cot A+\tan A)-2)(x+y-1)=0$
Comparing coefficients we get
$\cot A+\tan A=a+2$
$\dfrac {1}{\sin A \cos A}=a+2$

$2\sin A\cos A=\dfrac{2}{a+2}$
$=\sin 2A$
$=\sin 2\theta$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the two pair of lines $x^2-2mxy-y^2=0$ and $x^2-2nxy-y^2=0$ are such that one of them represents the bisectors of the angles between the other, then 

  1. $mn+1=0$
  2. $mn-1=0$
  3. $1/m+1/n=0$
  4. $1/m -1/n=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle bisectors of ax^2 + 2hxy + by^2 = 0 are given by (x^2-y^2)/(a-b) = xy/h. For x^2-2mxy-y^2=0 and x^2-2nxy-y^2=0, the bisectors of the first are (x^2-y^2)/(1-(-1)) = xy/(-m), which simplifies to x^2-y^2 = -2xy/m, or x^2 + (2/m)xy - y^2 = 0. Comparing this to the second equation, -2n = 2/m, so mn = -1, or mn+1=0.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the equation of the pair of straight lines passing through the point $(1, 1),$ one making an angle $\theta$ with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis is $x^{2}- (a + 2)xy + y^{2} + a(x + y -1) = 0, a \neq -2,$ then the value of $\sin 2\theta $ is

  1. $a -2$
  2. $a + 2$
  3. $\dfrac2{(a + 2)}$
  4. $ \dfrac2a$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equations of the given lines are $y -1 = \tan \theta (x -1) $ and $y -1 =\ cot \theta (x -1)$ 


so their joint equation is 

$[(y-1)-\tan \theta (x -1)][(y -1) -\cot \theta (x-1)] = 0$

$\Rightarrow (y -1)^{2} -(\tan \theta +\cot \theta) (x-l)(y -1) +(x-l)^{2}= 0$

$\Rightarrow x^{2} -(\tan \theta + cot \theta) xy + y^{2} + (\tan \theta+ \cot \theta -2) (x+y -1)=0$

Comparing with the given equation we get $\tan \theta + \cot \theta= a + 2$

$\displaystyle \Rightarrow \frac{1}{\sin \theta \cos\theta }= a + 2 \Rightarrow \sin 2\theta = \frac{2}{a+2}$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The absolute value of difference of the slope of the lines $\displaystyle x^{2}\left ( \sec ^{2}\theta -\sin ^{2}\theta  \right )-2xy\tan \theta +y^{2}\sin ^{2}\theta =0$ is

  1. $-2$
  2. $\dfrac{1}{2}$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given pair of lines
$x^2(\sec^2\theta-\sin^2\theta)-2xy\tan\theta+y^2\sin^2\theta=0$

$y^2\sin^2\theta-2xy\tan\theta+x^2(\sec^2\theta-\sin^2\theta)=0$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\tan^2\theta-4\sin^2\theta x^2(\sec^2\theta-\sin^2\theta)}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\tan^2\theta-4\tan^2\theta x^2+4x^2\sin^4\theta}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\sin^4\theta}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm2x\sin^2\theta}{2\sin^2\theta}$

$y=\dfrac{(\tan\theta\pm\sin^2\theta)}{\sin^2\theta}x$

On comparing above equation with $y=mx+c$ we get
$m=\dfrac{(\tan\theta\pm\sin^2\theta)}{\sin^2\theta}$

Here $m _{1}=\dfrac{(\tan\theta+\sin^2\theta)}{\sin^2\theta}$ and $m _{2}=\dfrac{(\tan\theta-\sin^2\theta)}{\sin^2\theta}$

$m _{1}-m _{2}=\dfrac{(\tan\theta+\sin^2\theta)}{\sin^2\theta}-\dfrac{(\tan\theta-\sin^2\theta)}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=\dfrac{\tan\theta+\sin^2\theta-\tan\theta+\sin^2\theta}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=\dfrac{2\sin^2\theta}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=2$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

lf the equation of the pair of straight lines passing through the point $(1,1 )$ , one making an angle ` $\theta$' with the postive direction of x-axis and the other making the same angle with the positive direction of y-axis is $x^{2}-(a+2)xy+y^{2}+a(x+y-1)=0$, $a\neq-2$, then the value of $\sin 2\theta$ is

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

Equations of the given lines are 
$y-1=\tan { \theta  } \left( x-1 \right) $ and $y-1=\cot { \theta  } \left( x-1 \right) $
Their combined equation is 
$\left( y-1-\tan { \theta  } \left( x-1 \right)  \right) \left( y-1-\cot { \theta  } \left( x-1 \right)  \right) =0\ \Rightarrow { x }^{ 2 }-\left( \tan { \theta  } +\cot { \theta  }  \right) xy+{ y }^{ 2 }+\left( \tan { \theta  } +\cot { \theta  } -2 \right) \left( x+y-1 \right) =0$
Comparing this with given equation we get
$\tan { \theta  } +\cot { \theta  } =a+2\ \Rightarrow \cfrac { 1 }{ \sin { \theta  } \cos { \theta  }  } =a+2\ \Rightarrow \sin { 2\theta  } =\cfrac { 2 }{ a+2 } $

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The locus of the mid point of the portion intercepted between the axes by the line $x{\,}cos\alpha+y{\,}sin{\,} \alpha=p$, where $p\inR$, is

  1. $x^2+y^2=\dfrac{4}{p^2}$
  2. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{4}{p^2}$
  3. $\dfrac{1}{x^2}-\dfrac{1}{y^2}=\dfrac{4}{p^2}$
  4. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{2}{p^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line x cos(alpha) + y sin(alpha) = p intercepts the axes at (p / cos(alpha), 0) and (0, p / sin(alpha)). The midpoint (h, k) of these intercepts is h = p / (2 cos(alpha)) and k = p / (2 sin(alpha)). Squaring and adding 1/h^2 + 1/k^2 yields 4/p^2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

I every points on the line $(a _{1}-a _{2})x+(b _{1}-b _{2}),y=c$ is equidistance from the points $(a _{1},b _{1})$  and $(a _{2},b _{2})$ then $2c=$  

  1. $a _{1}^{2}-b _{1}^{2}+a _{2}^{2}-b _{2}^{2}$
  2. $a _{1}^{2}+b _{1}^{2}+a _{2}^{2}+b _{2}^{2}$
  3. $a _{1}^{2}+b _{1}^{2}-a _{2}^{2}-b _{2}^{2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Any point on the given line is equidistant from (a1, b1) and (a2, b2), meaning it lies on the perpendicular bisector of the segment joining those two points. The equation of the perpendicular bisector can be found by equating the squared distances from a point (x, y) to both fixed points, which yields 2(a2 - a1)x + 2(b2 - b1)y = a2^2 + b2^2 - a1^2 - b1^2. Comparing this with the given line equation (a1 - a2)x + (b1 - b2)y = c, we can equate coefficients to find that 2c equals a1^2 + b1^2 - a2^2 - b2^2.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

Two system of rectangular axes have the same origin. If a plane cuts them at distances, $a$, $b$, $c$ and ${a} _{1}$,${b} _{1}$ , ${c} _{1}$ from the origin, then

  1. $\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$
  2. $\dfrac { 1 }{ { a }^{ 2 } } -\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$
  3. ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }+{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$
  4. ${ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }-{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the equation of the plane be
$\dfrac { x }{ a } +\dfrac { y }{ b } +\dfrac { z }{ c } =1$  and $\dfrac { x }{ a _1 } +\dfrac { y }{ b _1 } +\dfrac { z }{ c _1 } =1$
$ax+by+cz+d=0\quad perpendicular\quad distance\quad from\quad origin\quad is\quad \dfrac { \left| d \right|  }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }  } $
as they have the same origin their perpendicular distance is constant.
$\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  } =\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  }$
$\dfrac { 1 }{ { a _1}^{ 2 } } +\dfrac { 1 }{ { b _1 }^{ 2 } } +\dfrac { 1 }{ { c _1 }^{ 2 } } =\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } $


Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The line $2x+y =3$ cuts the ellipse $4x^2+y^2 =5$ at P and Q . If $\theta$ be the angle between the normals  at these point then $tan \theta$ =

  1. $1/2$
  2. $3/4$
  3. $3/5$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line 2x+y=3 intersects the ellipse 4x^2+y^2=5 at P(1, 1) and Q(1/2, 2). The slopes of the normals at these points are calculated using the derivative dy/dx = -4x/y. The slope of the normal at P(1, 1) is 1/4 and at Q(1/2, 2) is 1. The angle theta between them satisfies tan(theta) = |(1 - 1/4) / (1 + 1*1/4)| = (3/4) / (5/4) = 3/5.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between the planes $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $

  1. $\cos^{-1}\left(\displaystyle \frac{\bar{n _{1} }\cdot\bar{d} _{1}}{\left | \bar{d} _{1}\times \bar{d} _{2} \right |}\right)$
  2. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}.\bar{n} _{2}}{\left |\bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$
  3. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\bar{n} _{2}}{\bar{n} _{1}\times \bar{n} _{2} }\right)$
  4. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\cdot \left | \bar{d} _{2} \right |}{\left | \bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given planes are $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $ 

Angle between the planes is same as the angle between the normal vectors.
Hence the angle  $\theta=\cos^{-1}\left(\dfrac{\bar{n} _1.\bar{n} _2}{|\bar{n} _1||\bar{n} _2|}\right)$

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between two planes $\displaystyle r.n=q$ and $\displaystyle r.n'=q'$ is

  1. $\displaystyle \sin ^{-1}\left ( \frac{n.n'}{nn'} \right )$
  2. $\displaystyle \cos ^{-1}\left ( \frac{n.n'}{nn'} \right )$
  3. $\displaystyle \tan ^{-1}\left ( \frac{n.n'}{nn'} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given planes
$r\cdot n=q$------(1)
$r\cdot {n}'={q}'$-------(2)
Angle between two planes is between their normal vector 
$\left | n \right |\left | {n}' \right |cos\alpha=n \cdot {n}'$
$cos\alpha=\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |}$
$\alpha=\cos^{-1}(\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |})$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The line joining $A$ $\left( b\cos { \alpha ,\ b\sin { \alpha  }  }  \right)$ and $B$ $\left( a\cos { \beta ,\ a\sin { \beta  }  }  \right)$ is produced to the point $M$ $\left( x,y \right)$, so that $AM$ and $BM$ are in the ration $b:a$. Prove that
$x+y\ \tan { \left( \dfrac { \alpha +\beta  }{ 2 }  \right)  } =0$

  1. $-1$
  2. $0$
  3. $\sin (\alpha + \beta /2)$
  4. $\sin (\alpha - \beta /2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $\dfrac{AM}{BM}=\dfrac{b}{a}$
$\Rightarrow M$ divides $AB$ externally in the ratio $b:a$
$\Rightarrow x=\dfrac{ba\cos \beta-ab\cos \alpha}{b-a}$ and $y=\dfrac{ba\sin \beta-ab\sin \alpha}{b-a}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}$
$\cos \beta=\dfrac{1-\tan^2(\beta/2)}{1+\tan^2(\beta/2) }$, $\cos \alpha =\dfrac{1-\tan^2(\alpha /2)}{1+\tan^2(\alpha /2)}$, $\sin \beta=\dfrac{2\tan (\beta /2)}{1+\tan^2(\beta/2)}, \sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{1-\tan^2\dfrac{\beta}{2}}{1+\tan^2 \beta/2}-\dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2 \alpha/2}}{\dfrac{2\tan \beta/2}{1+\tan^2 \beta/2}-\dfrac{2\tan \alpha/2}{1+\tan^2 \alpha/2}}=\displaystyle \dfrac { 1+\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -1-\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }{ 2\tan { \dfrac { \beta  }{ 2 }  } +2\tan { \dfrac { \beta  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }  } $
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2\left( \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 } \beta /2 \right)  }{ 2\left( \tan  \beta /2-\tan  \dfrac { \alpha  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan { \beta /2 }  \right)  } =\dfrac { -\left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( \tan  \dfrac { \alpha  }{ 2 } +\tan  \dfrac { \beta  }{ 2 }  \right)  }{ \left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan  \dfrac { \beta  }{ 2 }  \right)  } $
$\Rightarrow x+y\dfrac{\tan \dfrac{\alpha}{2}+\tan \beta/2}{1-\tan \dfrac{\alpha}{2}\tan \dfrac{\beta}{2}}=0\Rightarrow x+y\tan \left(\dfrac{\alpha+\beta}{2}\right)=0$ Hence proved
Multiple choice

What is the bearing from point A to point B if the angle between the line connecting the two points and the north-south line is 120 degrees?

  1. N120E

  2. S120E

  3. N120W

  4. S120W
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The bearing from point A to point B is S120E because the angle between the line connecting the two points and the north-south line is 120 degrees east of south.

Multiple choice

What is the bearing from point A to point B if the angle between the line connecting the two points and the north-south line is 180 degrees?

  1. N180E

  2. S180E

  3. N180W

  4. S180W
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The bearing from point A to point B is S0W because the angle between the line connecting the two points and the north-south line is 0 degrees west of south.