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

In which of the following quadrants does the point (7, 3) lie?

  1. 2nd quadrant

  2. 3rd quadrant

  3. 1st quadrant

  4. 4th quadrant
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The point is (7, 3) $\leftrightarrow $ (x, y) Here, x > 0 and y > 0, and when x > 0 and y > 0, the point (x, y) lies in the first quadrant. Then, the point (7, 3) lies in the 1st quadrant.

Multiple choice

On which axis does the point (6, 0) lie?

  1. Negative y-axis

  2. Negative x-axis

  3. Positive y-axis

  4. Positive x-axis

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the point is (6, 0),             (6, 0) $\leftrightarrow $ (x, y) Here, x co-ordinate is 6, which is positive and y co-ordinate is 0. When y co-ordinate of a point is 0, then the point lies on x-axis. So, the point (6, 0) lies on positive x-axis.

Multiple choice

In which of the following quadrants does the point (4, – 8) lie?

  1. 2nd quadrant

  2. 3rd quadrant

  3. 1st quadrant

  4. 4th quadrant
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point is (4, – 8). Here, x co-ordinate is 4 > 0 and y co-ordinate is – 8 < 0. When x > 0 and y < 0, then the point (x, y) lies in the 4th quadrant. So, the point (4, – 8) lies in the fourth quadrant.                                                 OR The signs of co-ordinates of point are (+, –), which lies in the 4th quadrant. So, the point (4, – 8) lies in the 4th quadrant.

Multiple choice

In which of the following quadrants does the point (– 7, – 9) lie?

  1. 1st quadrant

  2. 2nd quadrant

  3. 3rd quadrant

  4. 4th quadrant

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The point is (– 7, – 9). Here, x co-ordinate is – 7, which is negative and y co-ordinate is – 9, which is also negative. When x < 0 and y < 0, then the point (x, y) lies in the third quadrant. So, the point (– 7, – 9) lies in the third quadrant.                                                 OR The signs of co-ordinates of the point are (–, –), which lies in the 3rd quadrant. So, the point (– 7, – 9) lies in the third quadrant.

Multiple choice combining transformations transformations vectors and transformations maths

The point $(4,3)$ is translated to the point $(3,1)$ and then axes are rotated through $30^{\mathrm{o}}$ about the origin, then the new position of the point is 

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

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

$x=\alpha +x^{1}\cos\theta -y^{1}\sin\theta  ; y=\beta +x^{1}\sin\theta +y^{1}\cos\theta $

$\Rightarrow 4=3+\dfrac{x^{1}\sqrt{3}}{2}-\dfrac{y^{1}}{2} ;3=1+\dfrac{x^{1}}{2}+\dfrac{y^{1}\sqrt{3}}{2}$

$\Rightarrow x^{1}\sqrt{3}-y^{1}=2$

$ x^{1}+y^{1}\sqrt{3}=4$

$\Rightarrow x=\dfrac{\sqrt{3}+2}{2}$

$y=\dfrac{2\sqrt{3}-1}{2}$

Multiple choice combining transformations transformations vectors and transformations maths

The point $A(2, 1)$ is translated parallel to the line $x- y = 3$ by a distance $4$ units. If the new position $A'$ is in third quadrant, then the coordinates of $A'$ are

  1. $(2 + 2 \sqrt{2}, 1 + 2\sqrt{2})$
  2. $(-2 + \sqrt{2}, -1 -2 \sqrt{2})$
  3. $(2 - 2 \sqrt{2}, 1 - 2 \sqrt{2})$

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

Since the point $A(2, 1)$ is translated parallel to $x-y =3$, therefore $AA'$ has the same slope as that of $x-y =3$. Therefore, $AA'$ passes through $(2, 1)$ and has the slope of $1$. 

Here, $\tan  \theta = 1 $
$\Rightarrow \cos \theta = \displaystyle\frac{1 }{\sqrt{2}}, \sin \theta = \displaystyle \frac {1 }{ \sqrt{2}}$
Using the parametric form of the line, the coordinates of $A'$ can be written as $\left(2\pm \displaystyle\frac {4}{\sqrt 2}, 1\pm \displaystyle\frac{4}{\sqrt 2}\right) $. 
Now, $A'$ is in the third quadrant. 
Hence, the coordinates of $A'$ are $(2- 2 \sqrt{2}, 1 - 2 \sqrt{2})$.

Multiple choice combining transformations transformations vectors and transformations maths

If the axes are rotated through an angle of ${30}^{o}$ in the anti-clockwise direction, the coordinates of point $(4,-2\sqrt{3})$ with respect to new axes are-

  1. $(2,\sqrt{3})$
  2. $(\sqrt{3}, -5)$
  3. $(2,3)$
  4. $(\sqrt{3},2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x=r\cos \alpha=4$

$y=r\sin\alpha=-2\sqrt3$
$x^1=r\cos(\alpha-30^o)$
$y^1=r\sin(\alpha-30^o)$
$x^1=r[\cos\alpha\cos 30^o+\sin\alpha\sin 30^o]$
$x^1=\dfrac{4\sqrt{3}}{2}-2\sqrt{3}\times \dfrac{1}{2}$
$x^1=2\sqrt{3}-\sqrt{3}$
$\therefore x^1=\sqrt{3}$
$y^1=r[\sin\alpha\cos 30^o-\cos \alpha\sin 30^o]$
      $=-2\sqrt{3}\times \dfrac{\sqrt{3}}{2}-4\times \dfrac{1}{2}$
$y^1=-5$
$\therefore \alpha(\sqrt{3}, -5)$

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

The points $(k, 2-2k)$, $(-k+1, 2k)$ and $(-4-k, 6-2k)$ are collinear for

  1. all values of k
  2. $k=-1$
  3. $k=1/2$

  4. no value of k.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

The given points are collinear if 


$\displaystyle \begin{vmatrix} k  & 2-2k  & 1 \ -k+1  & 2k  & 1 \ -4-k & 6-2k  & 1  \end{vmatrix}=0$

$\Rightarrow  
\begin{vmatrix} k  & 2-2k  & 1 \ -2k+1  & 4k-2  & 0 \ -4-2k & 4  & 0  \end{vmatrix}=0$       $ \left [ R _{2}\rightarrow R _{2}-R _{1}, R _{3}\rightarrow R _{3}-R _{1} \right ]$


$\Rightarrow 4(-2k+1)-(-4-2k)(4k-2)=0$

$\Rightarrow (1-2k)(4-8-4k)=0$

$\Rightarrow (1-2k)(k+1)=0$

$\Rightarrow k=-1 \ or \ k=1/2$