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 combining transformations transformations vectors and transformations maths

lf the axes are translated to the point $(-2, -3)$ , then the equation $\mathrm{x}^{2}+3\mathrm{y}^{2}+4\mathrm{x}+18\mathrm{y}+30=0$ transforms to

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=4$
  2. $\mathrm{X}^{2}+3\mathrm{Y}^{2}=1$
  3. $\mathrm{X}^{2}-\mathrm{Y}^{2}=4$
  4. $\mathrm{X}^{2} - 3 \mathrm{Y}^{2}=1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the point $(x,y)$ changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-2 , y=Y-3$
So, the equation transform to 
$(X-2)^2+3(Y-3)^2+4(X-2)+18(Y-3)+30=0$
$\Rightarrow X^2+3Y^2-1=0$
So, the transformed eqn is $X^2+3Y^2-1=0$

Multiple choice combining transformations transformations vectors and transformations maths

lf the origin is shifted to the point $(-1, 2)$ without changing the direction of axes, the equation ${x}^{2} -{y}^{2}+2{x}+4{y}=0$ becomes 

  1. ${X}^{2}+{Y}^{2}+3=0$
  2. ${X}^{2}+{Y}^{2}-3=0$
  3. ${X}^{2}-{Y}^{2}+3=0$
  4. ${X}^{2}-{Y}^{2}-3=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the point $(x,y)$ on the line changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-1 , y=Y+2$
So, the equation transform to 
$(X-1)^2-(Y+2)^2+2(X-1)+4(Y+2)=0$
$\Rightarrow X^2-Y^2+3=0$
So, the transformed eqn is $X^2-Y^2+3=0$

Multiple choice combining transformations transformations vectors and transformations maths

lf the axes are rotated through an angle $60^{\mathrm{o}}$, then the transformed equation of  $\mathrm{x}^{2}+\mathrm{y}^{2}=25$ is 

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=1$
  2. $\mathrm{X}^{2}+\mathrm{Y}^{2}=9$
  3. $\mathrm{X}^{2}+\mathrm{Y}^{2}=16$
  4. $\mathrm{X}^{2}+\mathrm{Y}^{2}=25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The radius of the circle is $5$. Despite rotating the reference frame, the origin still remains the same. Hence, the equation of the circle remains unchanged. 
Hence the new equation is:
$ X^2 + Y^2 =25 $

Multiple choice combining transformations transformations vectors and transformations maths

The transformed equation of $\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{y}\mathrm{s}\mathrm{i}\mathrm{n}\alpha = \mathrm{P}$ when the axes are rotated through an angle $\alpha$ is

  1. $\mathrm{X}=\mathrm{P}$
  2. $\mathrm{X}+\mathrm{P}=0$
  3. $\mathrm{Y}=\mathrm{P}$
  4. $\mathrm{Y}+\mathrm{P}=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By  rotation  of  axes,
$ x = x _{1} \cos\alpha -y _{1}\sin\alpha $
$ y = x _{1} \sin \alpha + y _{1} \cos \alpha $
$ x \cos  \alpha  + y \sin  \alpha = P$..........(given)
$ \Rightarrow  x _{1} (\cos^2 \alpha + \sin^{2}\alpha ) + y (\sin  \alpha  \cos  \alpha - \cos  \alpha  \sin  \alpha ) = P$..............(substitute the values of x and y) 
$x _{1} = P$ => X=P.

Multiple choice combining transformations transformations vectors and transformations maths

When axes are rotated by an angle of $135^{0}$, initial coordinates of the new coordinate $(4, -3)$ are             

  1. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
  2. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$
  3. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$
  4. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When axes are rotated through an angle $\theta $, then 

$x=X \cos\theta -Y \sin\theta$

$y=Y \cos\theta -X \sin\theta$

where(x,y) are initial coordinates and (X,Y) are new coordinates

$\therefore x=4\left(\dfrac{-1}{\sqrt{2}}\right)-\left(-3\right)\left(\dfrac{1}{\sqrt{2}}\right)=\dfrac{-1}{\sqrt{2}}$

$y=4\left(\dfrac{1}{\sqrt{2}}\right)-3\left(\dfrac{-1}{\sqrt{2}}\right)=\dfrac{7}{\sqrt{2}}$


$\therefore (x,y)=\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

Multiple choice combining transformations transformations vectors and transformations maths

If the axes are shifted to $(-2, -3)$ and rotated $\dfrac{\pi}{4}$ then Transformed equation of $2x^{2}+4xy-5y^{2}+20x-22y-14=0$ is 

  1. $X^{2}-14XY-7Y^{2}=2$
  2. $X^{2}-14XY-7Y^{2}=4$
  3. $X^{2}-14XY+7Y^{2}=2$
  4. $X^{2}+14XY+7Y^{2}=2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Shifting the origin to (-2, -3) and rotating by 45 degrees involves two steps: substitution of x = X' - 2, y = Y' - 3, followed by rotation. The resulting equation matches option A.

Multiple choice combining transformations transformations vectors and transformations maths

The transformed equation of $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$. When the coordinate axes are rotated through an angle of $45$, is

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

Since, the axes are rotated through an angle $45$, then we replace $\left( x,y \right) $ by
$\left( x\cos { 45 } -y\sin { 45 } ,x\sin { 45 } +y\cos { 45 }  \right) $
Given equation is $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$
Therefore, $ 3{ \left( \dfrac { x }{ \sqrt { 2 }  } -\dfrac { y }{ \sqrt { 2 }  }  \right)  }^{ 2 }+3\dfrac { x+y }{ \sqrt { 2 }  } +2\dfrac { x-y }{ \sqrt { 2 }  } \dfrac { x+y }{ \sqrt { 2 }  } =2$
$\Rightarrow \dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }+2xy \right) +\dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }-2xy \right) +\dfrac { 2 }{ 2 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$
$\Rightarrow  4{ x }^{ 2 }=2{ y }^{ 2 }=2$
$\Rightarrow  2{ x }^{ 2 }+{ y }^{ 2 }=1$

Multiple choice maths geometrical construction constructing perpendicular lines perpendicular to a line from an external point constructing an perpendicular line constructing a perpendicular bisector construction of a perpendicular bisector construction of penpendicual bisector set squares

When two lines are perpendicular to each other, the angle is said to be _______ angle.

  1. acute

  2. right

  3. obtuse

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

Two given lines are perpendicular means the angle between them is $90^o$, i.e. a right angle.

Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form

A line in the $xy$-plane passes through the origin and has a slope of $\dfrac{1}{7}$. Which of the following points lies on the line?

  1. $\left(0, 7\right)$
  2. $\left(1, 7\right)$
  3. $\left(7, 7\right)$
  4. $\left(14, 2\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If any straight line passes through origin, then it must of the form $y = mx$.


Now if the slope is $\dfrac{1}{7}$, then line will be $y = \dfrac{1}{7}x \ $ or $ \ 7y -x = 0$

We can see out of all the points only point $(14,2)$ satisfies the equation of the line. Hence Only $(14,2)$ lies on the line.

Correct option is $D$

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The pair of lines represented by $\displaystyle :3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and at right angles to each other, then value $ \left ( s \right )$ of $a$ is/are:

  1. $\displaystyle \:\frac{-3+\sqrt{17}}{2}$
  2. $\displaystyle \:\frac{-3-\sqrt{17}}{2}$
  3. $\displaystyle \:\frac{3+\sqrt{17}}{2}$
  4. $\displaystyle \:\frac{3-\sqrt{17}}{2}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The pair of lines represented by $\displaystyle :3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and at right angles to each other.
pair of lines given by $ax^2+2hxy+by^2=0$ are at right angles if $a+b=0$
$\Rightarrow 3a+a^2-2=0$
$\Rightarrow a^2+3a-2=0$
$\therefore a=\displaystyle\frac{-3\pm \sqrt{17}}{2}$
Hence, options A and B.

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

If one of the line given by the equation $a _{1}x^{2}+2h _{1}xy+b _{1}y^{2}=0$ coincides with one of the lines given by $a _{2}x^{2}+2h _{2}xy+b _{2}y^{2}=0$ and the other lines represented by them be perpendicular then $\dfrac {h _{1}a _{2}b _{2}}{b^{2}-a _{2}}\dfrac {h _{2}a _{1}b _{1}}{b _{1}-a _{1}}=\dfrac {1}{2}\sqrt {-a _{1}a _{2}b _{1}b _{2}}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a known identity in coordinate geometry regarding the intersection and perpendicularity of lines represented by homogeneous second-degree equations. The statement provided is a standard theorem result.

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

If $2x^{2}+3xy+my^{2}=0$ represents two real and mutually perpendicular lines then $m$ is

  1. any negative real number

  2. any positive real number
  3. $-2$

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

From above formula
$\tan90^{\circ}=\left | \dfrac{2\sqrt{(\frac{3}{2})^{2}+2m}}{2+m} \right |=\dfrac{1}{0}$
$m=-2$

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The product of the perpendiculars from origin to the pair of lines $ a x ^ { 2 } + 2 h x y + b y ^ { 2 } + 2 g x + 2 f y + c = 0 $ is

  1. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } + 4 h ^ { 2 } } }

    $
  2. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } - 4 h ^ { 2 } } }

    $
  3. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } + 4 h ^ { 2 } } }

    $
  4. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } - 4 h ^ { 2 } } }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The product of the perpendiculars from origin to the pair of lines $ a x ^ { 2 } + 2 h x y + b y ^ { 2 } + 2 g x + 2 f y + c = 0 $ is

  1. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } + 4 h ^ { 2 } } }

    $
  2. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } - 4 h ^ { 2 } } }

    $
  3. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } + 4 h ^ { 2 } } }

    $
  4. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } - 4 h ^ { 2 } } }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

If pair of lines $\displaystyle y^{2}+2hxy-9x^{2}=0$ and another pair of lines given by $\displaystyle ay^{2}+10xy+x^{2}=0$ have exactly one line common and other lines represented by them are perpendicular then

  1. a = 9; h = -4

  2. a + h = 6
  3. angle between the lines represented by first pair is $\displaystyle \frac{\pi }{4}$

  4. all of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $ \displaystyle m _{1} $ and $ \displaystyle m _{2} $ be the slope of lines of first pair


$ \displaystyle \Rightarrow  m _{1}$ and  $ \displaystyle \Rightarrow  m _{2}$  be the slope of lines of second pair

Compairing pair of lines (y-$ \displaystyle \Rightarrow  m _{1} x $ )  (y-$ \displaystyle \Rightarrow  m _{2} x $ ) with $ \displaystyle y^{2}+2hxy-9x^{2}=0 $

and pair of lines (y-$ \displaystyle \Rightarrow  m _{1} x $ ) $ \displaystyle y+\frac{1}{m _{2}}x $ with $ \displaystyle ay^{2}+10xy+x^{2}=0 $ 

we get $ \displaystyle m _{1}=-1\, and\, m _{2}=9\Rightarrow a=9\,and\, h=-4 $