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 lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If the lines $p _{1}x+q _{1}y=1,p _{2}x+q _{2}y=1 $ and $ p _{3}x+q _{3}y=1$ be concurrent, then the points $(p _{1},q _{1}),(p _{2},q _{2})$ and $(p _{3},q _{3})$ ,

  1. are collinear

  2. form an equilateral triangle

  3. form a scalene triangle

  4. form a right angled triangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$p _{1}x+q _{1}y=1,\ p _{2}x+q _{2}y =1\ p _{3}x+q _{3}y=1$
Given lines are concurrent
$\Rightarrow \begin{vmatrix} p _{1} & q _{1} & 1 \ p _{2} & q _{2}& 1 \ p _{3} & q _{3} & 1 \end{vmatrix}=0$

$\Rightarrow p _{1}(q _{2}-q _{3})-q _{1}(p _{2}-p _{3})+(p _{2}q _{3}-p _{3}q _{2})=0$

$\Rightarrow (p _{1}q _{2}-p _{2}q _{1})+(p _{2}q _{3}-p _{3}q _{2})+(p _{3}q _{1}-p _{1}q _{3})=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $(p _1, q _1),\; (p _2, q _2),\; (p _3,q _3)$
Since it is equal to zero, $(p _{1},q _{1}),(p _{2},q _{2}),(p _{3},q _{3})$ are collinear.

Multiple choice maths transformations midpoint of line segment mid point formula mid-point of a line segment

If $A(-2,5)$ and $B(3,2)$ are the points on a straight line. If ${AB}$ is extended to $'C'$ such that $AC=2BC$, then the co-ordinates of $'C'$ are ____

  1. $\left(\displaystyle\frac{1}{2}, \frac{3}{2}\right)$
  2. $\left(\displaystyle\frac{7}{2}, \frac{1}{2}\right)$
  3. $(8, -1)$
  4. $(-1, 8)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given points are $A(-2,5)$ and $B(3,2)$ 
$\dfrac{AC}{BC}=\dfrac{2}{1}$ , $C$ divides line segment $ { AB } $ externally.
If $A({ x } _{ 1 },{ y } _{ 1 })$ and $B({ x } _{ 2 },{ y } _{ 2 })$ be two end points of a line segment, then the coordinates of the point $P(x,y)$ that divides the line segment externally in the ratio $m:n$ is $\left( \dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n } ,\dfrac { m{ y } _{ 2 }-ny _{ 1 } }{ m-n }  \right) $.
Thus, the coordinates of $C$ are $\left( \dfrac { 2(3)-1(-2) }{ 2-1 } ,\dfrac { 2(2)-1(5) }{ 2-1 }  \right) $.
$=\left( \dfrac { 6+2 }{ 1 } ,\dfrac { 4-5 }{ 1 }  \right) $
$=(8,-1)$
Multiple choice maths transformations midpoint of line segment mid point formula mid-point of a line segment

Find the co-ordinates of a point C on AB produced such that $3AB = AC$, where $A = (3, 2)$ and $B = (-2, 4).$

  1. $(-12, 8)$
  2. $(8, 12)$
  3. $(12, 8)$
  4. $(-8, 12)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the above condition, we can observe that the point B  divides the line segment joining AC in 1;3 ratio, or $AC:AB=3:1$ or $AB:BC=2:1$. Let the coordinates of C be (x,y). Therefore,
$B(-2,4)=\left(\dfrac{1(x)+2(3)}{3},\dfrac{1(y)+2(2)}{3}\right)$
Or  $\dfrac{6+x}{3}=-2$ or $6+x=-6$ or $x=-12$. Similarly $\dfrac{4+y}{3}=4$ or $4+y=12$ or $y=8$.
Hence $C(x,y)=(-12,8)$

Multiple choice maths transformations midpoint of line segment mid point formula mid-point of a line segment

Find $x$ and $y$ if $(2,5)$ is the midpoint of points $(x,y)$ and $(-5,6)$.

  1. $x=4, y=9$
  2. $x=9, y=4$
  3. $x=-9, y=4$
  4. $x=9, y=-4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If the end points of a line segment is $(x,y)$ and $(-5,6)$ then the midpoint of the line segment has the coordinates:
$\left( \dfrac { x-5 }{ 2 } ,\dfrac { y+6 }{ 2 }  \right) =\left( 2,5 \right)$ ...(hint: using mid-point formula)
Now equating the points:  $\dfrac { x-5 }{ 2 } =2$ 
$\Rightarrow x-5=4$ 
$\Rightarrow x=9$
And, $\dfrac { y+6 }{ 2 } =5$ 
$\Rightarrow y+6=10$ 
$\Rightarrow y=4$

Hence, $x=9$ and $y=4$.

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

If $\overline { z } $ lies in the third quadrant then $z$ lies in the

  1. First quadrant

  2. Second quadrant

  3. Third quadrant

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

Fact:  $\overline z$ is image of $z$ in $x$-axis

So if $\overline z$ lies in third quadrant then $z$ will lie in second quadrant 

Multiple choice statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

A point $(a, b)$ is called a good point if both $a$ and $b$ are integers. Number of good points on the curve $xy$ $=$ $225$ are

  1. 20

  2. 18

  3. 16

  4. 14
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The order pair $(x, y)$ satisfying $xy=225$ are $(1, 225), (3, 75) (5, 45), (9, 25), (15, 15)$. Order can be changed in the first four pairs and both $x$ and $y$ can be negative also, so the no. of pairs $=2(2\times 4+1)=18$

Multiple choice maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

If P is (-3, 4) and My Mx (P)  shows the reflection of the  point P in the x"axis and then  the reflection of the image in the  y"axis, then My Mx (P) is

  1. (3, 4)

  2. (- 3, - 4)

  3. (`3, 4)

  4. (3, - 4)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Point P is (-3, 4). First, Mx (reflection in x-axis) changes sign of y-coordinate: (-3, 4) → (-3, -4). Then My (reflection in y-axis) changes sign of x-coordinate: (-3, -4) → (3, -4). The composition My Mx (P) results in (3, -4).

Multiple choice maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

When $ABCD$ is reflected over the $y$-axis to ${ A }^{ \prime  }{ B }^{ \prime  }{ C }^{ \prime  }{ D }^{ \prime  }$ , what can be the coordinates of ${ D }^{ \prime  }$ given that D lies in the $1^{st}$ quadrant?

  1. $\left( -12,1 \right) $
  2. $\left( -12,-1 \right) $
  3. $\left( 12,-1 \right) $
  4. $\left( 1,12 \right) $
  5. $\left( 1,-12 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

Let $ABCD$ be a square in which $A$ lies on the positive y-axis and $B$ lies on the positive x-axis. If $D$ is the point $(12, 17)$ the coordinates of $C$ are.

  1. $(17, 12)$
  2. $(17, 5)$
  3. $(14, 16)$
  4. $(15, 3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given A(0, yA) and B(xB, 0), and D(12, 17). Since it is a square, the vector AB must be perpendicular to AD and equal in length. Solving for the coordinates of the vertices based on the square properties leads to C(17, 12).

Multiple choice maths basic geometrical concepts and shapes construction of line segment and circle of given radius construction related to lines constructing line segment

Four distinct points $(2K , 3K), (1 , 0), (0 , 1)$ and $(0 , 0)$ lie on a circle when

  1. all values of $K$ are integral
  2. $0 < K < 1$
  3. $K < 0$
  4. For one values of $K$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let A = $(0,0)$ , B$(0,1)$ , C$(2k,3k)$ , D$(1,0)$

As we can see

$AD \perp AB$

$\angle A = 90$

$\angle C = 90$

$\implies m _{BC} \times m _{DC} = -1$

$\dfrac{3k - 1}{2k} \times {3k}{2k - 1} = -1$

$\implies k(9k - 3) = -(4k - 2)k$

$k = 0 , 9k + 4k = 2+ 3 \implies 13k = 5 \implies k = \dfrac{5}{13}$

But if $k = 0$, C will be $(0,0)$ which is A

$k = \dfrac{5}{13} $ only one point.

Multiple choice physics rigid body dynamics motion of rigid body rigid body equilibrium of a rigid body

Find the new coordinates of the point (3, 4), if the origin is shifted to the point (1, 3).

  1. (2, 1)

  2. (1, 2)

  3. (1, -2)

  4. (2, -1)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initially the origin was (0,0) and the point was (3,4). If the origin is now shifted to a new point (1,3), then the coordinates of the new point will be (X,Y)  with respect to the new origin(h=1, k=3).  Thus, (3,4) = (X+1, Y+3). Solving, we get, X=2 and Y=1

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

The line equally inclined to the coordinate axes and equidistant from points 
$A(1,-2)and B(3,4) is$

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

Lines equally inclined to axes have slopes 1 or -1. Equidistant from (1, -2) and (3, 4) means they pass through the midpoint (2, 1) or are parallel to the line segment AB. Calculating these yields the lines x+y=3 and x-y=1.