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 general knowledge math & puzzles

Angle between two parallel lines

  1. 90 degrees

  2. 180 degrees

  3. 290 degrees

  4. 0 degrees
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Parallel lines have the same direction and never intersect. The angle between them is conventionally measured as 0 degrees, representing the smallest angle between their direction vectors. The 180-degree option would indicate a straight line or opposite directions, not parallelism.

Multiple choice

On a graph paper, an insect moves from origin O and covers a distance of 4√3 in the direction of 60° east of north and reaches point A. Then, it turns and moves due south and reaches point B on x-axis and again it turns and reaches to the origin. What is the equation of line OB?

  1. x = 0

  2. x = 6

  3. y = 6

  4. y = 0

  5. y - x = 2√3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since point B is on x-axis, therefore the equation of line OB is y=0 as every point on x-axis has its y coordinate as 0. (Correct Answer)

Multiple choice

On a graph paper, an insect moves from origin O and covers a distance of 4√3 in the direction of 60° east of north and reaches point A. Then, it turns and moves due south and reaches point B on x-axis and again it turns and reaches to the origin. What is the equation of line AB?

  1. x = 6

  2. y = 6

  3. y = x - 6

  4. y = x + 6

  5. cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the insect turns and moves due south and cuts the x-axis at point B, the coordinates of point B are (6,0). (As x= 6 units from origin in the projection of A) Equation of the line is  x= 6 (Correct Answer)

Multiple choice

On a graph paper, an insect moves from origin O and covers a distance of 4√3 in the direction of 60° east of north and reaches point A. Then, it turns and moves due south and reaches point B on x-axis and again it turns and reaches to the origin. What is the equation of line OA?

  1. √3x - y = 0

  2. x - √3y = 0

  3. x + √3y = 0

  4. √3x + y = 0

  5. 4√3x - y = 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Coordinates of origin are (0,0). Since it moves at point A at a distance of 4√3 from origin in the direction of 60° east of north, hence the coordinates of point A are = (6, 2√3). Now, we are having the coordinates of O and A, we can easily find the equation of line. y-y1=m(x-x1), where m is the slope of the line. m = (y2-y1)/(x2-x1) y1=0, x1=0, y2=2√3, x2=6 m = (2√3/6) = 1/√3 Putting in the equation, we get x-√3y=0 (Correct Answer)  

Multiple choice
Directions: The following question consists of certain statements. You have to select the option representing the most logical sequencing of the given statements. A. Line X is perpendicular to line Y. B. Line Y is perpendicular to line Z. C. Line Z is perpendicular to line X. D. Line Z is parallel to line X. E. Line Z and line X is the same line. F. Angle between lines Z and X is π.

  1. FAC

  2. ABC

  3. ABF

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

Option (4) None of these Let us validate (4) by discussing options (1), (2) & (3) Option (1) FAC F à Angle between lines Z & X is π. A à Line X is perpendicular to line Y. Therefore, a suitable following statement should be that line Z is perpendicular to line Y, instead, C goes like Line Z is perpendicular to line X. It is neglected by sentence F that has marked angle between X & Z to be π. Option (2) ABC A à line X is perpendicular to line Y B à line Y is perpendicular to line Z C à line Z is perpendicular to line X These 3 sentences appear in a flow, each having exactly same sentence except the pair of lines related are different. This is not true. Option (3) ABF A à Line X is perpendicular to line Y B à Line Y is perpendicular to line Z F à Angle between lines Z & X is π Think twice before marking this option as correct answer. No matter this relation is correct mathematically but you cannot presuppose that perpendicular is a property denoting angle between two lines. Thus, it is incorrect.
Therefore, only option (4) is left valid for answer.

Multiple choice

In quadrantal bearing system, bearing of a line varies from

  1. 0º to 360º

  2. 0º to 180º

  3. 0º to 90º

  4. 0º N to 90º S
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 Quadrantel bearing, meaning value of fearing varies in a quarter from $0^0 to 90^0$ with direction and in the whole circle bearing, bearing varies from $0^0 to 360^0$

Multiple choice maths functions and graphs different forms of equation of a line

The direction with $+x-axis$ in which a straight line will be drawn through the point $\left(1,2\right)$ so that its point of intersection with the line $x+y=4$ may be at a distance $\sqrt { \dfrac { 2 }{ 3 }  }$ from the point $\left(1,2\right)$ can be:

  1. ${15}^{o}$
  2. ${30}^{o}$
  3. ${45}^{o}$
  4. ${75}^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the geometry: The line through (1,2) at angle θ to +x-axis meets x+y=4 at distance √(2/3). Using parametric form and distance formula gives θ = 15° as the valid solution. This involves setting up the line equation, finding intersection point, and applying distance constraint.

Multiple choice maths functions and graphs different forms of equation of a line
$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0 $ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0 $ or 
$ \displaystyle \begin{vmatrix}a&h  &g \\ h&b  &f \\ g&f  &c \end{vmatrix}=0 $ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

If the lines joining origin to the points of intersection of the line $\displaystyle x +y = 1$ with the curve $\displaystyle x^{2}+y^{2}+x-2y-m = 0$ are perpendicular to each other, then value of $m$ is

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

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$


$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( 1 \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( x+y \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x^{2}-xy-2y^{2} \right )-m\left ( x^{2}+y^{2}+2xy \right )=0$

$\displaystyle \Rightarrow  x^{2}\left ( 2-m \right )+y^{2}\left ( 1-2-m \right )+xy\left ( 1-2m \right )=0$      ......(*)

Now pair of straight lines given bye (*) will be $\perp$er if

$\displaystyle 2-m+1-2-m=0$ (Using $a+b=0$)

$\displaystyle \Rightarrow m=\frac{1}{2}$

Multiple choice maths functions and graphs different forms of equation of a line

The lines joining the origin to the point of intersection of $3x^2 + mxy - 4x + 1 = 0$ and $2x + y - 1= 0$  are at right angles. Then which of the following is/are possible value/s of $m?$

  1. $-4$
  2. $4$
  3. $7$
  4. $3$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $ x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

The angles between the lines $3 x + y - 7 = 0 \text { and } x + 2 y + 9 = 0$ is:

  1. $\dfrac { \pi } { 3 }$
  2. $\dfrac { \pi } { 6 }$
  3. $\dfrac { \pi } { 2 }$
  4. $\dfrac { \pi } { 4 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Slope of the line $y=-3x+7$ is $-3$
Slope of the line $2y=-x-9$ or $y=\dfrac{-1}{2}x-\dfrac{9}{2}$ is $-\dfrac{1}{2}$

$\therefore\,{m} _{1}=-3,\,{m} _{2}=-\dfrac{1}{2}$

Angle between the line $=\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{-3+\dfrac{1}{2}}{1+\left(-3\right)\left(-\dfrac{1}{2}\right)}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{1+\dfrac{3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{2+3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{5}{2}}\right|$

$\tan{\theta}=1$

$\therefore\,\theta=\dfrac{\pi}{4}$
Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

Find the angle between the lines 3x + 2y = 6 and x + y = 6

  1. 12$\displaystyle ^{\circ}$ 20'
  2. 11$\displaystyle ^{\circ}$ 19'
  3. 14$\displaystyle ^{\circ}$ 25'
  4. 13$\displaystyle ^{\circ}$ 06'
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Line  $3x+2y=6$

$=>2y=-3x+6$
$=>y=\frac{-3}{2}x+\frac{6}{2}$
$=>m'=\frac{-3}{2}x$
$\therefore tan\theta _1=\frac{-3}{2}$
$=>\theta _1=tan^{-1}(\frac{-3}{2})$
$=>\theta _1=-56.31$
Line is $x+y=6$
$=>y=x+6$
$=>m _2=-1$
$=>tan\theta _2=-1$
$=>\theta _2=tan^{-1}(-1)$
$=>\theta _2=-45$
Thus, ale between them $=(-45-56.31)$
                                          $=11.31^0$

Multiple choice combining transformations transformations vectors and transformations maths

When the axes are rotated through an angle $\dfrac{\pi}{6}$ , find the new coordinate for $(1,0)$

  1. $(\dfrac{\sqrt3}{2},\dfrac{-1}{2})$
  2. $(\dfrac{\sqrt4}{2},\dfrac{-1}{2})$
  3. $(\dfrac{\sqrt5}{2},\dfrac{-1}{2})$
  4. $(\dfrac{\sqrt3}{2},\dfrac{-1}{3})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\\ sin(\frac{\pi}{6})=(\frac{b}{1})\\\therefore b= (\frac{1}{2})\\ so new y-coordinate wil be  = (\frac{-1}{2})\\cos(\frac{\pi}{6})=(\frac{a}{1})\\\therefore a=(\frac{\sqrt3}{2})\\\therefore new x-coordinate =(\frac{\sqrt3}{2})$

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 $X^2-Y^2+3=0$