Mathematics · Quantitative Aptitude

Coordinate Geometry

221 Questions

Practice fundamental coordinate geometry questions covering reflections, collinear points, and loci. This collection helps in mastering XY plane plots, midpoints, and quadrant identification. It is highly beneficial for students preparing for quantitative aptitude tests, JEE, and state level engineering exams.

Point reflectionCollinear pointsXY plane lociFinding midpointsCoordinate quadrantsRhombus vertices

Coordinate Geometry 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 the coordinates of $\left(x,y\right)$ lie in first quadrant then 

  1. $x = 1, y = 7$
  2. $x = 6, y =-2$
  3. $x = -3, y=7$

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

In first quadrant $x$ and $y$ are positive. Hence $x$ takes the value $1$ and $y$ takes the value $7$

$\therefore \,x=1,\,y=7$

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 $O$ is origin and $C$ is the mid point of $A(2,-1)$ and $B(-4,3)$. Then value of $\overrightarrow { OC } $ is

  1. $\hat { i } +\hat { j } $
  2. $\hat { i } -\hat { j } $
  3. $-\hat { i } +\hat { j } $
  4. $-\hat { i } -\hat { j } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $C$ is the mid point of $A(2,-1)$ and $B(-4,3)$


$\therefore$ Coordinates of $C$ are
$\left( \cfrac { 2-4 }{ 2 } ,\cfrac { -1+3 }{ 2 }  \right) =\left( -1,1 \right) $
$\therefore \overrightarrow { OC } =-\hat { i } +\hat { j } $

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

Which statement is true?

  1. The x-axis is a vertical line
  2. The point $(-2 , 3)$ lies in the III quadrant

  3. Origin is the point of intersection of the x-axis and y-axis
  4. The point $(-3, -4)$ lies in the II quadrant
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

i) The x-axis ,a line parallel to it is called horizontal line

ii) The point (-2,3) lies in II Quadrant 
iii) Origin is the point of intersecting of x-axis and y-axis
iv) The point (-3,-4) lies in III quadrant .

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

Slope of the line $AB$ is $-\dfrac {4}{3}$. Co-ordinates of points $A$ and $B$ are $(x, -5)$ and $(-5, 3)$ respectively. What is the value of $x$

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

$\dfrac {y _{2} - y _{1}}{x _{2} - x _{1}} = \dfrac {-4}{3} =\dfrac{3+5}{-5-x}=\dfrac{-4}{3}$

$\Rightarrow 24 = 20 + 4x$
$x = 1$.

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

The coordinates of $A, B$ and $C$ are $(5, 5), (2, 1)$ and $(0, k)$ respectively. The value of $k$ that makes $\overline {AB} + \overline {BC}$ as small as possible is

  1. $3$
  2. $4\dfrac {1}{2}$
  3. $3\dfrac {6}{7}$
  4. $4\dfrac {5}{6}$
  5. $2\dfrac {1}{7}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The smallest possible value of $\overline {AC} + \overline {BC}$ is obtained when $C$ is the intersection of the y-axis, with the line that leads from $A$ to the mirror image the mirror being the y-axis) $B' : (-2, 1)$ of $B$. This is true because $\overline {CB'} = \overline {CB}$ and a straight line is the shortest path between two points. The line through and $B'$ given by
$y = \dfrac {5 - 1}{5 + 2} x + k = \dfrac {4}{7} x + k$.
To find $k$, we use the fact that the line goes through $A$:
$5 = \dfrac {4}{7} . 5 + k, k = 5 - \dfrac {20}{7} = \dfrac {15}{7} = 2\dfrac {1}{7}$;
$\therefore C$ has coordinates $(0, 2\dfrac {1}{7})$.

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

If the points $A ( 2,1,1 ) , B ( 0 , - 1,4 ) , C ( K , 3 , - 2 )$ are collinear then $K =$

  1. k=5

  2. k=4

  3. k=9

  4. k=10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For points A, B, and C to be collinear, the vectors AB and BC must be proportional. AB = (0-2, -1-1, 4-1) = (-2, -2, 3). BC = (K-0, 3-(-1), -2-4) = (K, 4, -6). Comparing components: -2/K = -2/4 = 3/-6. This gives -2/K = -1/2, so K = 4.

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

If the points $( 2,0 ) , ( 0,1 ) , ( 4,5 ) \text { and } ( 0 , c )$ are concyclic then the value of $c$ is 

  1. $1,\dfrac{13}{3}$
  2. $5,\dfrac { 14 } { 3 }$
  3. $5,\dfrac{15}{4}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider equation of circle to be $x^2+y^2+2gx+2fg+c=0$
Substituting given points
$4+4g+c=0$ ………….(i)
$1+2f+c=0$ ………..(ii)
$39+8g+10f+c=0$ ……….(iii)
solving them $g=-25/12$; $f=-8/3$
$c=13/3$
$x^2+y^2-\dfrac{25}{6}x-\dfrac{16}{3}y+\dfrac{13}{3}=0$ is equation
Substituting $(0, c)$
$c^2-\dfrac{16}{3}c+13/3=0$
$c=1, 13/3$.
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

If points $( - 7,5 ) \text { and } \left( \alpha , \alpha ^ { 2 } \right)$ lie on the opposite sides of the line $5 x - 6 y - 1 = 0$ then 

  1. $\alpha \in [ 0,1 ]$
  2. $a \in [ - 1,0 ]$
  3. $\alpha \in \left( \dfrac { 1 } { 3 } , \dfrac { 1 } { 2 } \right)$
  4. $\alpha \in [ 2,4 ]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(-7, 5)$ & $(\alpha, \alpha^2)$ lie on opposite side of $5x-6y-1=0$

$(-7, 5)\rightarrow 5x-6y-1=5(-7)-6(5)-1$

$=-35-30-1$

$=-66 < 0$

as $(\alpha,\alpha^2)$ should be on the opposite of $5x-6y-1=0$

$\Rightarrow (\alpha, \alpha^2)\rightarrow 5x-6y-1=5\alpha-6\alpha^2-1 > 0$

$\Rightarrow 6\alpha^2-5\alpha +1 < 0$

$\Rightarrow 6\alpha^2-2\alpha -3\alpha +1 < 0$

$\Rightarrow 2\alpha(3\alpha -1)-1(3\alpha -1) < 0$

$\Rightarrow (2\alpha -1)(3\alpha -1) < 0$

$\Rightarrow \dfrac{1}{3} < \alpha < \dfrac{1}{2}$.
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

If the three distinct points $\left( t,2at+{ at }^{ 3 } \right)$ for $i=1,2,3$are collinear then the sum of the abscissa of the  _________.

  1. -1

  2. 0

  3. 1

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The points are of the form (t, 2at + at^3). For these to be collinear, they must satisfy a linear equation. The sum of the abscissae (t1+t2+t3) for collinear points in this specific parametric form is 0.

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

If a point P from where line drawn cuts coordinate axes at A and B(with A on x-axis and B on y-axis) satisfies $\alpha\cdot \dfrac{x^2}{PB^2}+\beta\dfrac{y^2}{PA^2}=1$, then $\alpha +\beta$ is?

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

This is a property of the intercept form of a line. For a line segment divided by a point P, the relationship between the segments and the intercepts leads to alpha + beta = 1.

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

If the points $(k, 2 - 2k), (1 - k, 2k)$ and $(-k -4, 6 - 2k)$ be collinear the possible value(s) of $k$ is/are

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

Let the three points be,

$A=(k,2-2k)$
$B=(1-k,2k)$
$C=(-k-4,6-2k)$

If $A,B,C$ to be collinear, the area of the triangle formed by these three points must be $0$.

For these points to be collinear, we know that
$\left| \begin{matrix} k & 2-2k & 1 \ 1-k & 2k & 1 \ -k & 6-2k & 1 \end{matrix} \right| =0$

$k(2k-6+2k)-(2-2k)[(1-k)+k]+1[(1-k)(6-2k)+2k^2]=0$
$4k^2-6k-2+2k+6-8k+2k^2+2k^2=0$
$8k^2-12k+4=0$
$2k^2-3k+1=0$
$(2k-1)(k-1)=0$
$k=1,\dfrac{1}{2}$

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

If a point $P$ has coordinates $(3,4)$ in a coordinate system $X'OX\leftrightarrow  Y'OY$, and if $O$ has coordinates $(4,3)$ in another system ${X} {1}'{O} _{1}{X} _{1}\leftrightarrow  {Y} _{1}'{O} _{1}{Y} _{1}$ with $X'OX\parallel  {X} _{1}'{O} _{1}{X} _{1}$, then the coordinates of $P$ in the new system ${X} _{1}'{O} _{1}{X} _{1}\leftrightarrow  {Y} _{1}'{O} _{1}{Y} _{1}$ is _______________

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

Given the point P has coordinate (3,4).Similarly O has (4,3)

Given that $X'OX\leftrightarrow Y'OY\ { X } _{ 1 }^{ ' }{ O } _{ 1 }{ X } _{ 1 }\leftrightarrow { Y } _{ 1 }^{ ' }{ O } _{ 1 }{ Y } _{ 1 }\ \Longrightarrow { X }^{ ' }OX\parallel { X } _{ 1 }^{ ' }{ O } _{ 1 }{ X } _{ 1 }$
$\therefore$ Coordinate of P in new system ${ X } _{ 1 }^{ ' }O{ X } _{ 1 }\leftrightarrow { Y } _{ 1 }^{ ' }O{ Y } _{ 1 }\ =(3+4,)=(7,7)$