Mathematics · Quantitative Aptitude

Coordinate Geometry

204 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 direction cosines and direction ratios three dimensional geometry maths

For what value of $m$, the points $(3,5)$, $(m,6)$ and $\begin{pmatrix} \dfrac { 1 }{ 2 },\dfrac {15 }{ 2 } \end{pmatrix}$ are collinear?

  1. $9$
  2. $5$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. 
Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ $=$ $ $\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $
So, slope of the line joining $ (3,5) , (m,6) = $ Slope of the line joining $ (3,5) $ and $\left  (\dfrac {1}{2}, \dfrac {15}{2}\right ) $ 

Therefore, $ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $
$\Rightarrow  \dfrac { 1 }{ m - 3 } = -1 $

$\Rightarrow  m - 3 = -1 $

$\Rightarrow  m = 2 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(p,0)$, $(0,q)$ and $(1,1)$ are collinear, then $\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$ is equal to:

  1. $-1$
  2. $1$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 }

\right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ = $ $\dfrac { { y

} _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $

So, slope of the line joining $ (p,0) , (0,q) = $ Slope of the line joining

$ (0,q) $ and $ (1,1) $

$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $

$ - \dfrac { q }{ p } = 1 - q $

Dividing both sides by $q$,
$ - \dfrac { 1 }{ p } =  \dfrac { 1 }{ q } - 1 $

$ => \dfrac { 1 }{ p } +  \dfrac { 1 }{ q } = 1 $

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Coordinates of a point P are $(a, b)$ where $a$ is a root of the equation 

$x^{2}+x-42=0$ 
and $b$ is an integral root of the equation
$x^{2}+ax+a^{2}-37=0$. 
The coordinates of P can be

  1. $(6, 4)$
  2. $(-7, 4)$
  3. $(-7, 3)$
  4. $(6, -3)$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$a^{2}+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^{2}+ax+a^{2}-37=0$
For $a=-7,$ we have $x^{2}-7x+49-37=0$
$\Rightarrow x^{2}-7x+12=0\Rightarrow x=4, 3 , so, b=4$ or $3.$
So the coordinates of P can be $(-7, 4)$ or $(-7, 3)$, 

For $a=6,$ we have $x^{2}+6x-1=0$ which does not give an integral value, so $a\neq 6.$

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The point $\mathrm{A}(2,1)$ 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. $(2+2\sqrt{2},2+2\sqrt{2})$
  2. $(-2+\sqrt{2},-1-2\sqrt{2})$
  3. $(2-2\sqrt{2},1-2\sqrt{2})$
  4. $(-2-\sqrt{2},-1-2\sqrt{2})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ y = x-3$ $\Rightarrow m=1$ ; $tan \theta =1$
By parametrization we have $x=2 \pm r cos\theta $
$y=1 \pm r sin\theta $

$x =2 \pm 4\times \dfrac{1}{\sqrt{2}} ;\ y=1 \pm 4\times \dfrac{1}{\sqrt{2}}$.....(consider -ive sign for third quadrant)
$x=2-2\sqrt{2};\ y=1-2\sqrt{2}$ since they are in third quadrant.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

If a point $\mathrm{P}(4,3)$ is shifted by a distances $\sqrt{2}$ units parallel to the line $\mathrm{y}=\mathrm{x}$, then the coordinates of $\mathrm{P}$ in its new position are

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

shifted by $\sqrt{2}$ units $\Rightarrow r=\sqrt{2}$

for the line is tan $\theta =1$

$\therefore$ by parametric equations, we have

$x=x^{1}+r\ cos \theta$ ;  $y=y^{1}+r\ sin \theta$

$\Rightarrow x=4+\sqrt{2}\frac{1}{\sqrt{2}}$ ; $y=3+\sqrt{2}\frac{1}{\sqrt{2}}$

$x=5$  ;  $y=4$

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the point A$(1,2)$ by the line mirror y=x and the image of B by the line mirror $y=0$ is the point $\left(\alpha, \beta \right)$, then :

  1. $\alpha =1,\beta =-2$
  2. $\alpha =0,\beta =0$
  3. $\alpha =2,\beta =-1$

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

The image of (1, 2) in y=x is (2, 1). The image of (2, 1) in y=0 is (2, -1). Thus, alpha=2 and beta=-1.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Locus of the image of the point (2, 3) in the line (2x - 3y + 4) + k(x - 2y + 3) = 0, k $\in $ R, is a 

  1. straight line parallel to x-axis

  2. straight line parallel to y-axis
  3. Circle of radius $\sqrt { 2 } $

  4. circle of radius 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The family of lines (2x-3y+4) + k(x-2y+3) = 0 passes through a fixed point (intersection of the two lines). The locus of the image of a point reflected across a family of lines passing through a fixed point is a circle centered at that fixed point.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

If the line $\left (2\cos \theta+ 3\sin \theta\right)$ $x+(\left (3\cos \theta- 5\sin \theta\right)$ $y-\left (5\cos \theta- 2\sin \theta\right)=0$ passes through a fixed point $P$ for all values $\theta$ and $Q$ be the image of the point $P$ with the respect to the line $4x+6y-23=0$, then the distance of $Q$ from the origin is:

  1. $\dfrac {13}{5}$
  2. $\sqrt {5}$
  3. $5\sqrt {2}$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line equation can be rewritten as (2x+3y-5)cos(theta) + (3x-5y+2)sin(theta) = 0. This passes through the intersection of 2x+3y=5 and 3x-5y=-2. Solving this gives P(1, 1). Reflecting P across 4x+6y=23 gives Q, then calculate distance to origin.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the point $A(1,2)$ by the line mirror  $y=x$ is the 
Point B and the image of B by the line mirror $y=0$ is the point $(a,\beta )then:$ 

  1. $a = -2,\beta = - 1$
  2. $a = 0,\beta = 0$
  3. $a = 2,\beta = - 1$

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

The image of point A(1, 2) across the line y = x swaps its coordinates, giving point B(2, 1). Next, the image of B(2, 1) across the line y = 0 (the x-axis) reflects the y-coordinate, yielding (2, -1). Therefore, a = 2 and beta = -1.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The line segment joining $A\left( {3,\,\,0} \right),\,\,B\left( {5,\,\,2} \right)$ is rotated about a point A in anticlockwise sense through an angle $\displaystyle{\pi  \over 4}$ and B move to C. If a point D be the reflection of C in y-axis, then D=

  1. $\left( { - 3,\,2\sqrt 2 } \right)$
  2. $\left( {3,\,2\sqrt 2 } \right)$
  3. $\left( {3,\, - 2\sqrt 2 } \right)$
  4. $\left( {3,\,8\sqrt 2 } \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A(3, 0)$ and $B(5, 2)$
Slope of AB$=\dfrac{2-0}{5-3}=1$
Then, if $\theta$, is the angle made by AB, with positive direction of x-axis, we have $\tan\theta =1$
$\Rightarrow \theta =45^o$
Given, AB is rotated by $45^o$ to AC
Now AB$=\sqrt{(3-5)^2+(0-2)^2}=2\sqrt{2}$
So, coordinate of pr w$(3, 2\sqrt{2})$
Hence reflection of c in y-axis is $(-3, 2\sqrt{2})$.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Find the image of the point $\displaystyle \left ( -2, -7 \right )$ under the transformation
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right ).$

  1. $\displaystyle \left ( 12, -1 \right )$
  2. $\displaystyle \left ( -12, 1 \right )$
  3. $\displaystyle \left ( 2, 7 \right )$
  4. $\displaystyle \left ( -2, 7 \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given transformation is 
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right )$
So, the point $(-2,-7)$ $\rightarrow (-2-2(-7),-3(-2)-7)$
So, the image of the point $(-2,-7)$ under the given transformation is $ (12,-1)$

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to the line $L =$ $0$, then $L =$

  1. $6x - 4y -7 =0$
  2. $2x + 3y -5 =0$
  3. $3x - 2y + 5 =0$
  4. $3x - 2y + 10=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let points are $A(-2,6)$ & $B(4,2)$

$\therefore$    $L=0$ is $\bot $ to $AB$ & passes through the mid point slope $AB=\dfrac { -4 }{ 6 } =\dfrac { -2 }{ 3 } $
$\therefore$    slope of $L=0=3/2$
passes through $(1,4)$
$\therefore$    required equation is $\left( y-4 \right) =\frac { 3 }{ 2 } \left( x-1 \right) $
                                     $\Rightarrow 2y-8=3x-3$
                                     $\Rightarrow 3x-2y+5=0$        [C]

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the point $A(1, 2)$ by the line mirror $y=x$ is the point $B$ and the image of $B$ by the line mirror $y=0$ is the point $(\alpha, \beta)$, then?

  1. $\alpha =1, \beta =-2$
  2. $\alpha =0, \beta =0$
  3. $\alpha =2, \beta =-1$
  4. $\alpha =1, \beta =-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Point A(1,2) reflected in y=x gives B(2,1) (coordinates swap). B(2,1) reflected in y=0 gives (2,-1) (y-coordinate negated). Therefore alpha=2, beta=-1. This is identical to question 463562, testing the same two-step reflection concept.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The point $P(2, 1)$ is shifted by $\displaystyle 3\sqrt{2}$ parallel to the line $\displaystyle x+y=1,$ in the direction of increasing ordinate, to reach $Q$.The image of $Q$ by the line $\displaystyle x+y=1$ is

  1. $\displaystyle (5,-2)$
  2. $\displaystyle (-1,4)$
  3. $\displaystyle (3,-4)$
  4. $\displaystyle (-3,2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point $P(2,1)$ if shifted by $3\sqrt2$ parallel to the line $x+y =1$, in the direction of increasing ordinate, to reach point $Q(\alpha, \beta)$.


The equation of line produced from $P(2,1)$ to $Q(\alpha, \beta)$ parallel to line $x+y = 1$ in parametric form is,

$\Rightarrow \dfrac { \alpha -2}{cos \theta} = \dfrac{ \beta -1}{sin \theta} = -3\sqrt2$   .....$(1)$

Where $\theta$ is slope of line $PQ$, which is parallel to line $x +y = 1$

Hence Slope of $PQ$ $= -1$

$\Rightarrow\text{ if} \   tan \theta = -1 $

$\Rightarrow cos \theta = \dfrac {1}{\sqrt2}$

$\Rightarrow sin \theta = - \dfrac{1}{\sqrt2}$

Hence from equation $(1)$, $\alpha = -1, \beta = 4$ and $Q(-1,4)$

Now image of $Q(-1,4)$ in the line $x+y = 1$ is given by,

$\Rightarrow \dfrac {x - x _1 } {a} = \dfrac{y -y _1} {b} = \dfrac {-2 (ax _1 +by _1 +c)}{a^2 +b^2} $

$\Rightarrow \dfrac{x+1}{1} = \dfrac{y-4}{1} = \dfrac {-2 (-1 +4 -1)}{1^2 +1^2}$

Hence $x = -3$ and $y = 2$

Correct option is $D$.