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 hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The centre of the conic section $14x^2-4xy+11y^2-44x-58y+71=0$ is

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

Let $S\equiv 14x^2-4xy+11y^2-44x-58y+71$
To know centre of this hyperbola, $\cfrac{\partial S}{\partial x}=0$ and $\cfrac{\partial S}{\partial y}=0 $
$\Rightarrow \cfrac{\partial S}{\partial x}=28x-4y-44=0\Rightarrow 7x-y-11=0  ..(1)$
and $\cfrac{\partial S}{\partial y}=-4x+22y-58=0\Rightarrow 2x-11y+29=0  ..(2)$
Solving $(1)$ and $(2)$ we get required centre $(2,3)$
Hence, option 'A' is correct.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

A point $(\alpha, \beta)$ lies on a circle $x^2+y^2=1$, then locus of the point $(3\alpha +2\beta)$ is a$/$an.

  1. Straight line

  2. Ellipse

  3. Parabola

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Point will be $(3\alpha ,2\beta )$ not $( 3\alpha +2\beta )$
Now $ x^{2}+y^{2}=1 $
Radium is $1$ unit,hence parametric co - ordinate is 
$(\alpha ,\beta )= (1\cos\theta ,1\sin\theta )=(\cos\theta , \sin\theta )$
Hence
Point is $ (3\cos\theta ,2\sin\theta )$
Hence
$(x,y)= (3\cos\theta ,2\sin\theta )$
$x=3\cos\theta $
$ \Rightarrow \dfrac{x}{3}\cos\theta$    ...(i)
$ y=2\sin\theta $
$ \Rightarrow \dfrac{y}{2} = \sin \theta$   ...(ii)
$ (i)^{2} + (ii)^{2} $
$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} \cos^{2}\theta + \sin^{2} \theta $
$ \dfrac{x^{2}}{9}+ \dfrac{y^{2}}{4} = 1 $
which is equation of ellipse
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The locus of a point which moves such that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other two sides is

  1. Hyperbola

  2. A parabola

  3. An ellipse

  4. A ciircle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If the triangle is PQR, with $PQ = PR$, take P, Q, R to be the points $(0,b), (–a,0), (a,0)$ respectively.
The equation of the line PQ is 
$y-b=\dfrac{b}{+a}(x)$
$ay-ab=bx$
$bx-ay+ab=0$
and the equation of PR is
$y-b=\dfrac{b}{-a}(x)$
$-ay+ab=bx$
$bx+ay-ab=0$
and the equation of QR is
$y=\dfrac{0}{2a}(x)$
$y=0$
To find the locus of a point $X(h,k)$ which moves so that the square of its distance from QR is equal to the product of its distances from $PQ$ and $PR$. The distance from $X(h,k)$ to PQ is 

$d _{1}=\left | \dfrac{bh-ak+ab}{\sqrt{a^2+b^2}} \right |$
and the distance from $X(h,k)$ to PR is
$d _{2}=\left | \dfrac{bh+ak-ab}{\sqrt{a^2+b^2}} \right |$
The distance from $X(h,k)$ to QR is $|k|$
So according to question 
The distance of $X$ to QR=product of distances from $X$ to PQ and PR 
$k=d _{1}d _{2}$

$k=\dfrac{bh-(ak-ab)}{\sqrt{a^2+b^2}}\times\dfrac{bh+(ak-ab)}{\sqrt{a^2+b^2}}$

$k=\dfrac{b^2h^2-(a^2k^2+a^2b^2-2a^2kb)}{a^2+b^2}$

$a^2k+b^2k=b^2h^2-a^2k^2-a^2b^2+2a^2bk$
Putting $h=x,k=y$
$b^2x^2+(2a^2+b^2)y^2+2a^2by-a^2b^2=0$
Hence above equation represents the pair of straight lines

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The mid point of $(3,4)$ and $(1,-2)$

  1. (2,1)

  2. (1,2)

  3. (2,-1)

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

The points are $(3,4)$ and $(1,-2)$


The mid point of $(3,4)$ and $(1,-2)$ is given by 

$\left(\dfrac {x _1+x _2}2,\dfrac {y _1+y _2}2\right)\\\left(\dfrac{3+1}2,\dfrac {4-2}2\right)=(2,1)$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

If (-2, -4) is the midpoint of (6, -7) and (x, y) then the values of x and y are

  1. x = 2, y = 1

  2. x = -10, y = -1

  3. x = 10, y = -1

  4. x = -8 , y = -1

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

Since, $(-2, -4)$ is the midpoint of $(6, -7)$ and $(x,y).$
$\Rightarrow \dfrac{x+6}2=-2\Rightarrow x=-10$
and $ \dfrac{y-7}2=-4\Rightarrow y=-1$
Option D is correct.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The coordinates of points $P(-2, 2), Q(3, 2) $ and $R(3, -2)$ are the vertices of a rectangle $PQRS$`. What are the coordinates of S? 

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

Mid-point of $PR=\left(\cfrac{-2+3}2,\cfrac{2-2}2\right)=\left(\cfrac12,0\right)$

Let coordinate of $S=(x,y)$
Mid-point of $QS=\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$
Mid-point of $PR=$Mid-point of $QS$
$\Rightarrow\left(\cfrac12,0\right)=$$\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$
We have $\cfrac12=\cfrac{3+x}2\Rightarrow x=-2$ and $0=\cfrac{2+y}2\Rightarrow y=-2$
Coordinate of $S=(-2,-2)$
Hence, B is the correct option.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

If $O(0,0)$ and $P(-8,0)$ then co-ordinates of its midpoint are________.

  1. $(-4,0)$
  2. $(4,0)$
  3. $(0,-4)$
  4. $(0,0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$O(0,0) $ and $P(-8,0)$ are given points.


The coordinates of the midpoint of 

$\overline{OP}=\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2}\right)$

         $= \left( \dfrac {0-8}{2},\dfrac {0-0}{2}\right)$

         $=(-4,0)$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The point on $X-axis$ equidistant from $(2,3)$and $(1,5)$ is

  1. $\left( \dfrac { -13 }{ 2 } ,0 \right) $
  2. $\left( \dfrac { 13 }{ 2 } ,0 \right) $
  3. $(13,0)$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the point be (x, 0). Equidistance means (x-2)^2 + (0-3)^2 = (x-1)^2 + (0-5)^2. Expanding: x^2 - 4x + 4 + 9 = x^2 - 2x + 1 + 25. Simplifying: -4x + 13 = -2x + 26, so -2x = 13, x = -13/2. The point is (-13/2, 0).

Multiple choice maths constructions mid-point formula midpoints division of a line segment

I every points on the line $(a _{1}-a _{2})x+(b _{1}-b _{2}),y=c$ is equidistance from the points $(a _{1},b _{1})$  and $(a _{2},b _{2})$ then $2c=$  

  1. $a _{1}^{2}-b _{1}^{2}+a _{2}^{2}-b _{2}^{2}$
  2. $a _{1}^{2}+b _{1}^{2}+a _{2}^{2}+b _{2}^{2}$
  3. $a _{1}^{2}+b _{1}^{2}-a _{2}^{2}-b _{2}^{2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer