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 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 business economics and quantitative methods linear regression aspects of correlation scatter graphs and correlation correlation

The angle between the two lines will be wider when the correlation between two variable is___________.

  1. less

  2. more

  3. equal

  4. very high
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The larger the angle between the two regression lines, less the degree of correlation between the two variables and similarly, the small the angle between two lines, the larger would be the degree of correlation between them.

Multiple choice business economics and quantitative methods correlation analysis aspects of correlation scatter graphs and correlation linear regression

When r=0, the lines of regression will____________.

  1. intersect each other at $90^o$

  2. be positive slope

  3. be negative slope

  4. be convex to the origin
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the correlation coefficient r is 0, there is no linear relationship between the variables. Consequently, the two regression lines are perpendicular to each other, intersecting at a 90-degree angle.

Multiple choice business economics and quantitative methods correlation analysis aspects of correlation scatter graphs and correlation linear regression

When r = +1, the lines of regression will_____________.

  1. be positive slope

  2. move upwards from left to right

  3. be negative slope

  4. both (A) and (B)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When coefficient of correlation is +1, it is referred to as perfect positive correlation and the two lines of regression coincide and it has a positive slope.

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\,  } }& \, { 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}$

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 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. $