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

The product of the perpendiculars from origin to the pair of lines ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$ is


  1. $\frac { \left| c \right| }{ \sqrt { \left( { a+b } \right) ^{ 2 } } +{ 4h }^{ 2 } } $
  2. $\frac { \left| c \right| }{ \sqrt { \left( a+b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
  3. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
  4. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
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

A line is at distance of $4$ units from origin and having both intercepts positive. If the perpendicular from the origin to this line makes an angle of ${60}^{o}$ with the line $x+y=0$ Then the equation of the line is

  1. $\left( \sqrt { 3 } +1 \right) x+\left( \sqrt { 3 } +2 \right) y=y=8\sqrt { 2 } $
  2. $\left( \sqrt { 3 } -1 \right) x+\left( \sqrt { 3 } +1 \right) y=y=8\sqrt { 2 } $
  3. $\left( \sqrt { 3 } +1 \right) x-\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $
  4. $\left( \sqrt { 3 } +2 \right) x+\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is at distance 4 from the origin. Using the normal form x cos(theta) + y sin(theta) = 4, and the condition that the normal makes 60 degrees with x+y=0 (which has a normal vector (1,1) at 45 degrees), the angle of the normal is 45 +/- 60 degrees. Calculating the intercepts and checking the positive intercept condition leads to the correct equation.

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

A line passes through (3, 0) The slope of the line for which its intercept between y = x - 2 and y = -x + 2 subtends a right angle at the origin may be

  1. $\displaystyle \sqrt{2}$
  2. $\displaystyle -\sqrt{2}$
  3. $\displaystyle \frac{1}{\sqrt{3}}$
  4. $\displaystyle -\frac{1}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given line 
$y=x-2\Rightarrow x-y-2=0----(1)$
$y=-x+2\Rightarrow x+y-2=0----(2)$
On multiplying eq (1) and (2)
$(x-y-2)(x+y-2)=0$
$x^2+4-4x-y^2=0$
$x^2-y^2-4x+4=0---(3)$
Equation of line from point $(3,0)$ with slope m 
$y=mx-3m$
$1=\dfrac{mx-y}{3m}$
From eq (3)
$x^2-y^2-4x\left ( \dfrac{mx-y}{3m} \right )+4\left ( \dfrac{mx-y}{3m} \right )^2=0$

$x^2-y^2-\left ( \dfrac{4mx^2-4xy}{3m} \right )+4\left ( \dfrac{m^2x^2+y^2-2mxy}{9m^2} \right )=0$

$9m^2x^2-9m^2y^2-12m^2x^2+12mxy+4m^2x^2+4y^2-8mxy=0$

$(m^2)x^2+(-9m^2+4)y^2+4mxy=0$

Since line subtends right angle 
$m^2-9m^2+4=0$
$8m^2=4$
$m^2=\dfrac{1}{2}$
$m=\pm\dfrac{1}{\sqrt{2}}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Let $P _{1},\ P _{2},\ P _{3}$ be the perpendicular distances between pair of parallel lines represented by $x^{2}-3x-4=0$, $y^{2}-5y+6=0$, $4x^{2}+20xy+25y^{2}=0$ respectively then 

  1. $P _{3} < P _{2} < P _{1}$
  2. $P _{3} < P _{1} < P _{2}$
  3. $P _{2} < P _{1} < P _{3}$
  4. $P _{1} < P _{2} < P _{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2}-3x-4=0$
$(x-4)(x+1)=0$

$x=4$ and $x=-1$
Hence the perpendicular distance between these two lines 
$P _{1}=4-(-1)=5$.

$y^{2}-5y+6=0$
$(y-2)(y-3)=0$
$y=2$ and $y=3$
Hence perpendicular distance between these lines is 
$P _{2}=3-2=1$
Thus $P _{2}<P _{1}$ 

$4x^{2}+20xy+25y^{2}=0$
$x=\dfrac{-20y\pm\sqrt{400y^{2}-400y^{2}}}{8}$

$x=\dfrac{-20y}{8}$
Or 
$8x+20y=0$
$2x+5y=0$
Since we get a single line 
$P _{3}=0$
Therefore 
$P _{3}<P _{2}<P _{1}$

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

A straight lines moves such that the algebraic sum of the perpendicular drawn to it from two fixed points is equal to 2k than, the straight line always touches a fixed circle of radius.

  1. 2k
  2. $ \frac{k}{2} $

  3. k

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the line be x cos(alpha) + y sin(alpha) = p. The sum of perpendiculars from (x1, y1) and (x2, y2) is |x1 cos(alpha) + y1 sin(alpha) - p| + |x2 cos(alpha) + y2 sin(alpha) - p| = 2k. This is a standard locus problem where the line touches a circle.

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

The product of the perepndiculars drawn from the point $\left(x _1,y _1\right)$ on the lines $ax^2+2hxy+by^2=0$ is

  1. $\displaystyle \frac { a{ { x } _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  2. $\displaystyle \frac { \left| a{ { x } _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } \right| }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  3. $\displaystyle \frac { a{ { x } _{ 1 } }^{ 2 }-2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  4. $\displaystyle \frac { \left| a{ { x } _{ 1 } }^{ 2 }-2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } \right| }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\displaystyle y={ m } _{ 1 }x$ and $\displaystyle y={ m } _{ 2 }x$ be the two lines given by $\displaystyle{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$ so that

$\displaystyle{ m } _{ 1 }+{ m } _{ 2 }=\frac { -2h }{ b } $ and $\displaystyle{ m } _{ 1 }{ m } _{ 2 }=\frac { a }{ b } $   ...(1)
The product of the perpendiculars drawn from $\displaystyle\left( { x } _{ 1, }{ y } _{ 1 } \right) $ on these lines

$\displaystyle=\frac { \left| { y } _{ 1 }-{ m } _{ 1 }{ x } _{ 1 } \right|  }{ \sqrt { 1+{ { m } _{ 1 } }^{ 2 } }  } .\frac { \left| { y } _{ 1 }-{ m } _{ 2 }{ x } _{ 1 } \right|  }{ \sqrt { 1+{ { m } _{ 2 } }^{ 2 } }  } $

$\displaystyle =\frac { { { y } _{ 1 } }^{ 2 }-\left( { m } _{ 1 }+{ m } _{ 2 } \right) { x } _{ 1 }{ y } _{ 1 }+{ m } _{ 1 }{ m } _{ 2 }{ { x }^{ 2 } } _{ 1 } }{ \sqrt { 1+{ { m }^{ 2 } } _{ 1 }+{ { m } _{ 2 } }^{ 2 }+{ { m } _{ 1 } }^{ 2 }{ { m } _{ 2 } }^{ 2 } }  } $

$=$$\displaystyle\dfrac { \left| { { { y } _{ 1 } }^{ 2 }-\left( { m } _{ 1 }+{ m } _{ 2 } \right) { x } _{ 1 }{ y } _{ 1 }+{ m } _{ 1 }{ m } _{ 2 }{ { { { x } _{ 1 } }^{ 2 } } } } \right|  }{ \sqrt { 1+{ \left( { m } _{ 1 }+{ m } _{ 2 } \right)  }^{ 2 }-2{ m } _{ 1 }{ m } _{ 2 }+{ { m } _{ 1 } }^{ 2 }{ { m } _{ 2 } }^{ 2 } }  } $

$\displaystyle=\dfrac { \left| { { y } _{ 1 } }^{ 2 }+\dfrac { 2h{ x } _{ 1 }y _1 }{ b } +\dfrac { a{ { x } _{ 1 } }^{ 2 } }{ b }  \right|  }{ \sqrt { 1+\dfrac { 4{ h }^{ 2 } }{ { b }^{ 2 } } -\dfrac { 2a }{ b } +\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }  }  } $     (using(1) )

$\displaystyle=\frac { \left| { { a }x _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+{ { by } _{ 1 } }^{ 2 } \right|  }{ \sqrt { { \left( a-b \right)  }^{ 2 }+4{ h }^{ 2 } }  } $

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
A Correct answer
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 } } $