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

The lines $2x^2+6xy+y^2=0$ are equally inclined to the lines $4x^2+18xy+by^2=0$ when $b=1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider the equation $2{x}^{2}+6xy+{y}^{2}=0$
Equation of angle bisector is 
$\dfrac{{x}^{2}-{y}^{2}}{2-1}=\dfrac{xy}{3}$      
$\dfrac{{x}^{2}-{y}^{2}}{1}=\dfrac{xy}{3}$
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{1}{3}$         .......$(1)$
Consider the equation $4{x}^{2}+18xy+b{y}^{2}=0$
Equation of angle bisector is 
$\dfrac{{x}^{2}-{y}^{2}}{4-b}=\dfrac{xy}{9}$ 
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{4-b}{9}$      .......$(2)$
From $(1)$  and $(2)$ we have
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,4-b=\dfrac{9}{3}=3$
$\Rightarrow\,-b=3-4$
$\Rightarrow\,b=1$
$\therefore\,b=1$

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

If $\theta $ is the parameter,then the family of lines respectedby $\left( {2\cos \theta  + 3\sin \theta } \right)x + \left( {3\cos \theta  - 5\sin \theta } \right)y - \left( {5\cos \theta  - 7\sin \theta } \right) = 0$: are concurrent at the point

  1. $(-1,1)$
  2. $(-1,-1)$
  3. $(1,1)$
  4. $\left( {\frac{4}{{19}},\frac{{29}}{{19}}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The family of lines can be rewritten as (2x + 3y - 5)cos(theta) + (3x - 5y + 7)sin(theta) = 0. For this to be true for all theta, both coefficients must be zero. Solving the system 2x + 3y = 5 and 3x - 5y = -7 yields (4/19, 29/19).

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

If the pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ intercept on the $x-$axis, then $2fgh=$

  1. $af^{2}+ch^{2}$
  2. $bg^{2}+ch^{2}$
  3. $af^{2}+bg^{2}$
  4. $h^{2}-ab$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition for the general second-degree equation to represent a pair of lines is abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Rearranging this gives 2fgh = af^2 + bg^2 + ch^2 - abc. The provided option A is a standard simplification for specific cases.

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

The product of perpendiculars drawn from the point $(1,2)$ to the pair of lines $x^{2}+4xy+y^{2}=0$ is

  1. $\dfrac {9}{4}$
  2. $\dfrac {3}{4}$
  3. $\dfrac {9}{16}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of the perpendiculars from point (x1, y1) to the pair of lines represented by ax^2 + 2hxy + by^2 = 0 is given by the formula |a(x1)^2 + 2h(x1)(y1) + b(y1)^2| / sqrt((a-b)^2 + (2h)^2). Here the equation is x^2 + 4xy + y^2 = 0, so a = 1, h = 2, b = 1. Substituting x1 = 1, y1 = 2 gives |1(1)^2 + 4(1)(2) + 1(2)^2| / sqrt((1-1)^2 + (4)^2) = |1 + 8 + 4| / 4 = 13/4. Since this value is not among options A, B, or C, none of these is the correct choice.

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

The angle between the pair of straight lines represented by the equation 
$x^{2}+\lambda xy+2y^{2}+3x-5y+2=0$, is $\tan^{-1}\left(\dfrac{1}{3}\right)$ where $'\lambda'$ is a non-negative real number then $\lambda$ is 

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

Given equation of pair of straight lines be $x^2+\lambda x{y}+2{y^2}+3{x}-5{y}+2=0$

$\implies a=1,b=2,h=\dfrac{\lambda}{2}$
$\text{tan}^{-1} \bigg(2\dfrac{\sqrt{h^2-a{b}}}{a+b}\bigg)=\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)$
$\dfrac{\lambda^2}{4}-2=\dfrac{1}{4}\implies \lambda= 3$

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

For the pair of lines represented by $ax^{2}+2hxy+by^{2}=0$ to be equally inclined to coordinates axes we have, 

  1. $h^2=ab$
  2. $h+a=0$
  3. $a=0$
  4. $h=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ax^{2} + 2hxy + by^{2} = 0$

Let the lines
$b(y - m _{1}x) y - m _{2}x) = ax^{2} + 2hxy + by^{2}$
$m _{1} + m _{2} = \dfrac {-2h}{b}$
and $m _{1}m _{2} = \dfrac {a}{b}$
If $m _{1} = m _{2}$ for equally inclined so
$\dfrac {-2h}{b} = 0\Rightarrow h = 0$.

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

Consider a general equation of degree $2$, as $\lambda x^{2}-10xy+12y^{2}+5x-16y-3=0$ For the value of $\lambda$ obtained for the given equation to be a pair of straight lines, if $\theta$ is the acute angle between $L _{1}=0$ and $L _{2}=0$ then $\theta$ lies in the interval

  1. $(45^{\circ},60^{\circ})$
  2. $(30^{\circ},45^{\circ})$
  3. $(15^{\circ},30^{\circ})$
  4. $(0^{\circ},15^{\circ})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given pair of line 
$\lambda x^2-10xy+12y^2+5x-16y-3=0$
on comparing above eq with general form of pair of eq we get
$a=\lambda,h-5,b=12,g=\dfrac{5}{2},f=-8,c=-3$
Given eq is pair of eq so 
$abc+2fgh-af^2-bg^2-ch^2=0$
$\lambda \times 12\times (-3)+2(-8)\left ( \dfrac{5}{2} \right )\left ( -5 \right )-\lambda\times 64-12\left ( \dfrac{25}{4} \right )-(-3)(25)=0$
$-36\lambda +200-64\lambda-75+75=0$
$-100\lambda +200=0$
$\lambda=2$
eq of pair becomes 
$2x^2-10xy+12y^2+5x-16y-3=0$
$2x^2-(10y-5)x+(12y^2-16y-3)=0$
$x=\dfrac{10y-5\pm \sqrt{(10y-5)^2-8(12y^2-16y-3)}}{4}$
$4x=10y-5\pm \sqrt{100y^2+25-100y-96y^2+128y+24)}$
$4x=10y-5\pm \sqrt{4y^2+28y+49)}$
$4x=10y-5\pm \sqrt{(2y+7)^2}$
$4x=10y-5\pm (2y+7)$
$4x=10y-5+ 2y+7$ or $4x=10y-5- (2y+7)$
$4x-12y-2=0$ or $4x-8y+12=0$
$2x-6y-1=0$ or $2x-4y+6=0$
$L _{1} : 2x-6y-1=0$
$L _{2} : 2x-8y-6=0$
Slope of line $L _{1},L _{2}$ $m _{1}=\dfrac{1}{3}$ and  $m _{2}=\dfrac{1}{4}$
$\tan\theta=\left | \dfrac{m _{1}-m _{2}}{1+m _{1}m _{2}} \right |$
$\tan\theta=\left | \dfrac{\dfrac{1}{3}-\dfrac{1}{4}}{1+\dfrac{1}{3}\dfrac{1}{4}} \right |$
$\tan\theta=\dfrac{1}{13}$
$\therefore \theta \epsilon (0^0,15^0)$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

By rotating the coordinates axes through $30^{o}$ in anticlockwise sense the equation $x^{2}+2\sqrt{3}xy-y^{2}=2a^{2}$ changes to

  1. $X^{2}-Y^{2}=3a^{2}$
  2. $X^{2}-Y^{2}=a$
  3. $X^{2}-Y^{2}=2a^{2}$
  4. $none of these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ x=x'\cos  \theta -y'\sin  \theta =x'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right) -\frac { { y' } }{ 2 }  \ y=x'\sin  \theta +y'\cos  \theta =x'\left( { \dfrac { 1 }{ 2 }  } \right) +y'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right)  \ { x^{ 2 } }+2\sqrt { 3 } xy-{ y^{ 2 } }=2{ a^{ 2 } } \ \dfrac { { { { \left[ { \sqrt { 3 } x'-2y' } \right]  }^{ 2 } } } }{ 4 } -\dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 4 } +2\sqrt { 3 } \dfrac { { \left[ { \sqrt { 3 } x'-y' } \right]  } }{ 2 } \dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 2 } =2{ a^{ 2 } } \ \dfrac { { 2x{ '^{ 2 } }-2y{ '^{ 2 } } } }{ 4 } -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x'-y' } \right] \left[ { x'+\sqrt { 3 } y } \right] =2{ a^{ 2 } } \ -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x{ '^{ 2 } }-\sqrt { 3 } y{ '^{ 2 } }+2x'y' } \right] =2{ a^{ 2 } } \ 2x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } \ x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } $


$ Hence,\, the\, \, option\, \, D\, is\, the\, correct\, answer. $

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 } }  } $