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 mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

If a line passes through the point $P(1, 2)$ makes an angle of $45^o$ with the x-axis and meets the line $x+2y-7=0$ in Q, then PQ equals?

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

The line through (1, 2) with angle 45 degrees is y - 2 = 1(x - 1), or y = x + 1. Intersection with x + 2y - 7 = 0: x + 2(x + 1) - 7 = 0 => 3x = 5 => x = 5/3, y = 8/3. Distance PQ = sqrt((5/3 - 1)^2 + (8/3 - 2)^2) = sqrt((2/3)^2 + (2/3)^2) = sqrt(8/9) = 2*sqrt(2)/3.

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The acute angle between the lines $x-y=0$ and $y=0$ is

  1. $30^{\circ}$
  2. $45^{\circ}$
  3. $60^{\circ}$
  4. $75^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 


$x-y=0$

$y=0$ represents $x-axis$ 

So the slope of line $x-y=0$ is $-\dfrac 1{-1}=1$

$\implies \tan \theta =1$

$\tan \theta =\tan \dfrac \pi 4$

$\theta =\dfrac \pi 4$

$\theta =45^{\circ}$

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

To remove Xy term from the second degree equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$, the coordinates axes are rotated through an angle q, then q equals.

  1. $\pi/2$
  2. $\pi/4$
  3. $\pi/8$
  4. $\pi/8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To eliminate the xy term from a second-degree equation of the form ax^2 + hxy + by^2, the required rotation angle theta is given by tan(2theta) = h / (a - b). Here h = 8 and a = b = 5, leading to tan(2theta) = 8 / 0 = infinity, which means 2theta = pi/2 and theta = pi/4.

Multiple choice maths construction of angles identifying angles acute and obtuse angles types of angle

What is the acute angle between the lines y = 3x + 2 and y = 4x + 9?

  1. 4.4$\displaystyle ^{\circ}$
  2. 28.3$\displaystyle ^{\circ}$
  3. 5.2$\displaystyle ^{\circ}$
  4. 18.6$\displaystyle ^{\circ}$

  5. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$y=3x+2$

$m _1=3$
$=>tan\theta _1=3$
$=>\theta _1=tan^{-1}(3)$
$=>\theta _1=71.57^{0}$
Again,
$y=4x+9$
$=>m _2=4$
$=>tan\theta _2=4$
$=>\theta _2=tan^{1}(4)$
$=>\theta _2=75.96^{0}$
$\therefore$ Angle between the lines
$=\theta _2-\theta _1$
$=75.96-71.56$
$=4.4^{0}$

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

If the straight line through the point $P(3,4)$ makes an angle $\cfrac{\pi}{6}$ with the x-axis and meets the line $3x+5y+1=0$ at $Q$, the length $PQ$ is

  1. $\dfrac {132}{12\sqrt {3}+5}$
  2. $\dfrac {132}{12\sqrt {3}-5}$
  3. $\dfrac {132}{5\sqrt {3}+12}$
  4. $\dfrac {132}{5\sqrt {3}-12}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of straight line passing through $P(3,4)$ is $y=\tan \dfrac{\pi}{6}{x}+(4-3\tan \dfrac{\pi}{6})\implies y=\dfrac{x}{\sqrt{3}}+4-\sqrt{3}$

The point of intersection will be $\bigg(\dfrac{55-57\sqrt{3}}{5+3\sqrt{3}},\dfrac{-10+3\sqrt{3}}{5+3\sqrt{3}}\bigg)$
Length will be $30(5-3\sqrt{3})$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

A line is drawn from $P(x _1 , y _1)$ in the direction $\theta$ with the X - axis, to meet $ax + by + c = 0$ at $Q$. Then length $PQ$ is equal to :

  1. $\dfrac{|ax _1 + by _1 + c|}{\sqrt{(a^2 + b^2)}}$
  2. $\left|\dfrac{ax _1 + by _1 + c}{a \, cos \theta + b \, sin \theta} \right|$
  3. $\dfrac{bx _1 + ay _1 + c}{a cos \theta + b sin \theta}$
  4. $ - \dfrac{ax _1 + by _1 + c}{a sin \theta + b cos \theta}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of a line drawn through a point $P\left( { x } _{ 1 }{ y } _{ 1 } \right) $ at an angle $\theta $ with the $x-axis$
$\dfrac { x-{ x } _{ 1 } }{ \cos\theta  } =\dfrac { y-{ y } _{ 1 } }{ \sin\theta  } \quad \longrightarrow \left( 1 \right) $
$Q$ is a point the above line and also lies on the line $ax+by+cz=0$
Say $\left| PQ \right| =r$ and the coordinates of $Q$ are $\left( h,k \right) $
Then,
$h={ x } _{ 1 }+r\cos\theta $
$k={ y } _{ 1 }+r\sin\theta $
$\because$   $Q$ lies on $ax+by+cz=0$
$\therefore$   $ah+bk+c=0$
$\Rightarrow a\left( { x } _{ 1 }+r\cos\theta  \right) +b\left( { y } _{ 1 }+r\sin\theta  \right) +c=0$
$\Rightarrow { ax } _{ 1 }+{ by } _{ 1 }+c+r\left( a\cos\theta +b\sin\theta  \right) =0$
$\Rightarrow \quad r=\dfrac { -\left( { ax } _{ 1 }+{ by } _{ 1 }+c \right)  }{ a\cos\theta +b\sin\theta  } $
$\because$   $'r'$ is the magnitude of length of line segment $PQ$, it cannot be negative.
So, $r=\left| \dfrac { { ax } _{ 1 }+{ by } _{ 1 }+c }{ a\cos\theta +b\sin\theta  }  \right| $
Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

If a line which passes through the point $A(0,\,1,\,2)$ and makes angle $\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{2}$ with $x,\,y,\,&\,z$ axes respectively. The line meets the plane $x+y+z=0$ at point $B$. The length $\sqrt{2}AB$ is equal to

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

The D.C. of the line are $\begin{pmatrix}\displaystyle\frac{1}{\sqrt{2}},\,\displaystyle\frac{1}{\sqrt{2}},\,0\end{pmatrix}$
any point on the line at a distance $\lambda$ from $A(0,\,1,\,2)$
is $\begin{pmatrix}0+\displaystyle\frac{\lambda}{\sqrt{2}},\,1+\displaystyle\frac{\lambda}{\sqrt{2}},\,2+\lambda.0\end{pmatrix}$
which lies on $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

A line with positive direction cosines passes through the point $\displaystyle P\left ( 2,-1,2 \right )$ and makes equal angles with the coordinates axis. The line meet the plane $\displaystyle 2x+y+z=9$ at ponit $Q$.
The length of the line segment $PQ$ equals.

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

Equation of a line through the point $\displaystyle P\left ( 2,-1,2 \right )$, equally inclined to the axes is given by,
$\displaystyle \frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=k$ (say)
Any point on the line is $Q ( k+2,k-1, k+2  )$ which lies on the plane $\displaystyle 2x+y+z=9$
 $\Rightarrow \displaystyle 2\left ( k+2 \right )+k-1+k+2=9 \Rightarrow k=1$,
For this values of $k$, the coordinates of $Q$ are $\displaystyle \left ( 3,0,3 \right )$.
So, $\displaystyle PQ=\sqrt{\left ( 3-2 \right)^2+\left ( 0+1 \right )^2+\left ( 3-2 \right )^2}=\sqrt{3}$

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

If a line with direction ratio $2:2:1$ intersects the line $\dfrac {x-7}{3}=\dfrac {y-5}{2}=\dfrac {z-3}{1}$ and $\dfrac {x-1}{2}=\dfrac {y+1}{4}=\dfrac {z+1}{3}$ at $A$ and $B$ then $AB=$

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

$\begin{array}{l} \frac { { x-7 } }{ 3 } =\frac { { y-5 } }{ 2 } =\frac { { z-3 } }{ 1 } \, \, \, \, \, \, \, \, Let\, \, A=\left( { 3{ r _{ 1 } }+7,\, \, 2{ r _{ 1 } }+5,\, \, { r _{ 1 } }+3 } \right)  \ \frac { { x-1 } }{ 2 } =\frac { { y+1 } }{ 4 } =\frac { { z+1 } }{ 3 } \, \, \, \, \, and,\, Let\, B=\left( { 2{ r _{ 2 } }+1,\, \, 4{ r _{ 2 } }-1,\, \, 3{ r _{ 2 } }-1 } \right)  \ Direction\, ratios\, of\, AB=\left( { 2{ r _{ 2 } }-3{ r _{ 1 } }-6,\, \, \, 4{ r _{ 2 } }-2{ r _{ 1 } }-6,\, \, \, 3{ r _{ 2 } }-{ r _{ 1 } }-4 } \right)  \ \frac { { 2{ r _{ 2 } }-3{ r _{ 1 } }-6 } }{ 2 } =\frac { { 4{ r _{ 2 } }-2{ r _{ 1 } }-6 } }{ 2 } =\frac { { 3{ r _{ 2 } }-{ r _{ 1 } }-4 } }{ 1 }  \ { r _{ 1 } }+2{ r _{ 2 } }=0\, \, \, \, \therefore { r _{ 1 } }=-2 \ and, \ 4{ r _{ 1 } }-2{ r _{ 1 } }-6=6{ r _{ 2 } }-2{ r _{ 1 } }-8 \ 2=2{ r _{ 2 } } \ \therefore { r _{ 2 } }=1 \ A\equiv \left( { 1,\, 1,\, 1 } \right) \, \, \, \, \, B=\left( { 3,\, 3,\, 2 } \right)  \ AB=\sqrt { 4+4+1 }  \ =3 \ Hence,\, Option\, D\, is\, the\, correct\, answer. \end{array}$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The axis of the conic $\displaystyle x^{2}+4y-6x+17=0$ is

  1. $\displaystyle x=5 $
  2. $\displaystyle y=5 $
  3. $\displaystyle x=3 $
  4. $\displaystyle x=-3 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2 + 4y -6x +17 =0$

$\Rightarrow x^2 -6x = -4y-17$
$\Rightarrow x^2 - 6x + 9 = -4y + 8$
$\Rightarrow (x-3)^2 = -4(y-2)$
Let $X = x-3$ and $Y = y-2$, then we get
$X^2 = -4Y$
Comparing the above equation with the standard equation of the parabola, we get that the axis of the parabola is given by $X=0$
Hence, axis of the given parabola will be $x-3 = 0\Rightarrow x =3$
Option $C$ is correct.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Find the point of intersection and the inclination of the two lines $Ax+By=A+B$ and $A(x-y)+B(x + y)=2B$.

  1. $(1,1); 45^0$
  2. $(1,2), 60^0$
  3. $(2,1), 75^0$

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

$Ax+By=A+B$    ........(i)

$  A(x-y)+B(x+y)=2B$

$\Rightarrow  (A+B)x+(B-A)y=2B$    .......(ii)

$\Rightarrow  (A+B)x=2B-(B-A)y\\ \Rightarrow x=\dfrac { 2B-(B-A)y }{ A+B } $

Substituting $x$ in (i), we get

$A\left( \dfrac { 2B-(B-A)y }{ A+B }  \right) +By=A+B$

$\Rightarrow \dfrac { 2AB-ABy+{ A }^{ 2 }y+{ B }^{ 2 }y+ABy }{ A+B } =A+B$

$\Rightarrow 2AB-ABy+{ B }^{ 2 }y+{ A }^{ 2 }y+ABy={ A }^{ 2 }+{ B }^{ 2 }+2AB$ 

$\Rightarrow (A^{2}+{ B }^{ 2 })y=A^{2}+{ B }^{ 2 }$

$\Rightarrow y=1 $

Substituting $y$ in $(i)$

$\Rightarrow  Ax+B\left( 1  \right) =A+B\\ \Rightarrow Ax+B=A+B\\ \Rightarrow Ax=A\\ \Rightarrow x=1 $

So, the point of intersection is $(1,1) $.

Slope of (i), ${ m } _{ 1 }=-\dfrac { A }{ B } $.

Slope of (ii), ${ m } _{ 2 }=-\dfrac { (A+B) }{ B-A } =\dfrac { A+B }{ A-B } $

$\tan { \theta  } =\dfrac { { m } _{ 1 }-{ m } _{ 2 } }{ 1+{ m } _{ 1 }{ m } _{ 2 } } \\ \Rightarrow \tan { \theta  } =\dfrac { -\dfrac{A}{B}-\dfrac { A+B }{ A-B }  }{ 1-\dfrac{A}{B}\times\dfrac { A+B }{ A-B }  } $

$ \tan { \theta  } =-\dfrac { \left\{ \dfrac { { A }^{ 2 }+AB-AB+{ B }^{ 2 } }{ B(A-B) }  \right\}  }{ \left\{ \dfrac { { -B }^{ 2 }+AB-{ A }^{ 2 }-AB }{ B(A-B) }  \right\}  } \\ \tan { \theta  } =-\dfrac { { A }^{ 2 }+{ B }^{ 2 } }{ { -(A }^{ 2 }+{ B }^{ 2 }) } =1$

$\Rightarrow \theta=45^{o}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The pair of lines $6{ x }^{ 2 }+7xy+\lambda { y }^{ 2 }=0\left( \lambda \neq -6 \right) $ forms a right angled triangle with $x+3y+4=0$ then $\lambda=$

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

Given line is $L: x+3y+4=0$


$\implies  y=-\dfrac{1}{3}(x+4)$

Slope of this line is $m=\dfrac{-1}{3}$

Now, $6x^2+7xy+\lambda y^2=0$ $(\lambda\neq 6)$

$x^2+\dfrac{7}{6}xy+\dfrac{\lambda}{6}y^2=0$

$\implies (x+ay)(x+by)=0$

$\implies x+ay=0$ and $x+by=0$ are the two equations with 

$a+b=\dfrac{7}{6}$    and $ab=\dfrac{\lambda}{6}$

Slope of these lines are $m _1=\dfrac{-1}{a}$ and $m _2=\dfrac{-1}{b}$

Now, $m _1m _2=\dfrac{1}{ab}=\dfrac{\lambda}{6}\neq -1$  since $\lambda\neq -6$

Hence the lines $x+ay=0$ and $x+by=0$ are not prependicular.

From these two only one is normal to $L$.

Let $x+ay$ is normal to $L$.

$\implies m _1m=-1$

$\implies \dfrac{1}{3a}=-1$

$\implies a=\dfrac{-1}{3}$

Now, $a+b=\dfrac{7}{6}\implies b=\dfrac{7}{6}-\dfrac{-1}{3}$

$\implies b=\dfrac{3}{2}$

Now, $ab=\dfrac{\lambda}{6}$

$\implies \lambda=6ab=6\times \dfrac{-1}{3}\times \dfrac{3}{2}$

$\implies \lambda=-3$

Answer-(B)

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Assertion(A): The angle between the asymptotes of $3x^{2}-y^{2}=3$ is $120^{\circ}$
Reason(R): The angle between the asymptotes of $x^{2}-y^{2}=a^{2}$ is $90^{\circ}$

  1. Both A and R are true and R is the correct

    explanation of A.

  2. Both A and R are true but R is not correct

    explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Asymptotes of a hyperbola is given by $y=\pm \frac { x }{ a } $


So for hyperbola $3{ x }^{ 2 }-{ y }^{ 2 }=3$
The asymptotes make an angle of ${ 60 }^{ o }$ and  ${ 120 }^{ o }$ with x-axis which means they make an angle of ${ 60 }^{ o } $ among themself.

Now for hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={a}^{2}$

The asymptotes makes and angle of ${45}^{o}$ and  ${135}^{0}$ which means ${90}^{o}$ among themself.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The following lines are $\hat { r } =\left( \hat { i } +\hat { j }  \right) +\lambda \left( \hat { i } +2\hat { j } -\hat { k }  \right) +\mu \left( -\hat { i } +\hat { j } -\hat { 2k }  \right) $

  1. collinear

  2. skew-lines

  3. co-planar lines

  4. parallel lines
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Condition for three lines $\vec { { r } _{ 1 } } $ , $\vec { { r } _{ 2 } } $ , and $\vec { { r } _{ 3 } } $ to be collinear is:

$\vec { { r } _{ 1 } } +\lambda \vec { { r } _{ 2 } } +\vec { { \mu r } _{ 3 } } =0$
where $\vec { { r } _{ 1 } } =\left( \vec { i } +\vec { j }  \right) $
$\vec { { r } _{ 2 } } =\left( \vec { i } +2\vec { j } -\vec { k }  \right) $
$\vec { { r } _{ 3 } } =\left( -\vec { i } +\vec { j } -2\vec { k }  \right) $
and $\lambda $ and $\mu $ are scalars
Hence, the answer is collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the lines $x=1+a,y=-3-\lambda a,z=1+\lambda a$ and $x=\cfrac { b }{ 2 } ,y=1+b,z=2-b$ are coplanar, then $\lambda$ is equal to

  1. $-3$
  2. $2$
  3. $1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The given lines are $\cfrac { x-1 }{ 1 } =\cfrac { y+3 }{ -\lambda  } =\cfrac { z-1 }{ \lambda  } =\left( a \right) $ and $\cfrac { x-0 }{ 1/2 } =\cfrac { y-1 }{ 1 } =\cfrac { z-2 }{ -1 } =(b)$
$\therefore$ coplanarity, we must have
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda  & \lambda  \\ 1/2 & 1 & -1 \end{vmatrix}=0$
$\Rightarrow -1\left( \lambda -\lambda  \right) -4\left( -1-\cfrac { \lambda  }{ 2 }  \right) +\left( 1+\cfrac { \lambda  }{ 2 }  \right) =0$
$4+2\lambda +1+\cfrac { \lambda  }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda  }{ 2 } =0\quad \therefore \lambda =-2$