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

The point (4, 1) undergoes the following transformation successively.
(i)reflection about the line y=x
(ii)translation through a distance 2 units along the positive direction of x-axes.
(iii)rotation through an angle ${ \pi  }/{ 4 }$ about the origin in the anticlockwise direction.
(iv) reflection about x=0
The final position of the given point is

  1. $(1\sqrt { 2 } ,7/2)$
  2. $(1/2,7\sqrt { 2 } )$
  3. $(1\sqrt { 2 } ,7/\sqrt { 2 } )$
  4. $(1/2,7/2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Follow the steps: (4, 1) -> (1, 4) -> (3, 4) -> rotate by 45 deg -> reflection about x=0.

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

The point $A(4, 1)$ undergoes following transformations successively:
(i) reflection about line $y=x$
(ii) translation through a distance of $3$ units in the positive direction of x-axis.
(iii) rotation through an angle $105^o$ in anti-clockwise direction about origin O.
Then the final position of point A is?

  1. $\left(\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$
  2. $(-2, 7\sqrt{2})$
  3. $\left(-\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$
  4. $(-2\sqrt{6}, 2\sqrt{2})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The point A(4, 1) undergoes following transformations successively
(i) reflection about line y = x
(ii) translation through a distance of 2 units in the positive direction of x axis
(iii) rotation through an angle $\displaystyle \pi/4 $ in anti clockwise direction about origin O
Then the final position of point A is

  1. $\displaystyle \left ( \frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}} \right ) $
  2. $\displaystyle \left ( -2,7\sqrt{2} \right )$
  3. $\displaystyle \left ( -\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}} \right )$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given Point $A(4,1)$
$(i) $ reflection about $y=x$
Hence putting coordinates of point A in given eq $y=4,x=1$
The point becomes $A(1,4)$

$(ii)$ translation through distance of 2 units in positive X axis 
$A(1+2,4)\Rightarrow A(3,4)$

$(iii)$ rotation of point through and angle $\dfrac{\pi}{4}$ in anticlockwise about origin O
After rotation Point will be $A(r\cos(\alpha+\dfrac{pi}{4}),r\sin(\alpha+\dfrac{pi}{4}))$
Converting into polar form 
$r=\sqrt{3^2+4^2}=5$
$\tan\alpha=\dfrac{4}{3}$
Hence $\cos\alpha=\dfrac{3}{5}$
$\sin\alpha=\dfrac{4}{5}$
$\cos(\alpha+\dfrac{\pi}{4})=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{3}{5}\dfrac{1}{\sqrt{2}}-\dfrac{4}{5}\dfrac{1}{\sqrt{2}}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{-1}{5\sqrt{2}}$

$\sin(\alpha+\dfrac{\pi}{4})=\sin\alpha\cos\dfrac{\pi}{4}+\cos\alpha\sin\dfrac{\pi}{4}$
$\sin(\alpha+\dfrac{\pi}{4})=\dfrac{4}{5}\dfrac{1}{\sqrt{2}}+\dfrac{3}{5}\dfrac{1}{\sqrt{2}}$
$\sin(\alpha+\dfrac{\pi}{4})=\dfrac{7}{5\sqrt{2}}$
$A\left(5\times \dfrac{-1}{5\sqrt{2}},5\times \dfrac{7}{5\sqrt{2}}\right)$
$A\left(-\dfrac{1}{\sqrt{2}}, \dfrac{7}{\sqrt{2}}\right)$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The co-ordinates of the point of reflection of the origin $(0, 0)$ in the line $4x -2y - 5 = 0$ is

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

Let $(h,k)$ be the point of reflection.
Line joining $(0,0)$ and $(h,k)$ is perpendicular to $4x-2y-5=0$
Product of slopes of these lines is $-1$
$\dfrac{k}h*2=-1=>k=-\dfrac{h}2$
Midpoint of $(0,0)$ and $(h,k)$ i.e, $(\dfrac{h}{2},\dfrac{k}{2})$ lies on $4x-2y-5=0$
Therefore $4\dfrac{h}2-2\dfrac{k}2-5=0=>2h-k-5=0$ but $k=-\dfrac{h}2$
$2h+\dfrac{h}2-5=0=>h=2 $ and $k=-1$
Point of reflection is $(2,-1)$.

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


The point $(4, 1)$ undergoes the following three transformations successively
i) Reflection about the line $\mathrm{y}=\mathrm{x}$
ii) Transformation through a distance of $2$ units along the $+\mathrm{v}\mathrm{e}$ direction of the x-axis
iii) Rotation through an angle $\displaystyle \frac{\pi}{4}$ about the origin in the anticlockwise direction. The final position of the point is given by the co-ordinates 

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

Under the transformation reflection about $y = x$ the new point is $(1,4)$
Under the transformation of shifting it $+2$ units along x-axis the new point is $(1+2,4) = (3,4)$
Under the transformation of rotation of $\dfrac{\pi}{4}$
$y= -X \sin  \theta +y\cos\ \theta$

$X=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{7}{\sqrt{2}}$

$y=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}$

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

The image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ is

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

Image of any Point $(\alpha , \beta)$ with respect to line $ax+by +c = 0$ is given by equation,


$\Rightarrow \dfrac{x -\alpha}{a} = \dfrac{y - \beta}{b} = \dfrac{-2(a\alpha + b \beta + c )}{a^2 + b^2}$  ....$(1)$

Now the given point is $(3,8)$ and the given line is $x + 3y - 7 = 0$

Putting the values in equation  $(1)$, we get,

$\Rightarrow \dfrac{x -3}{1} = \dfrac{y - 8} {3} = \dfrac{ -2(1.3 + 3.8 -7 )}{1^2 + 3^2} = -4$

Hence $\Rightarrow x= -1$ and $y = -4$

Correct option is $A$.

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

 The point $(4, 1)$ undergoes the following three transformations successively
(a) Reflection about the line $y = x$

(b) Transformation through a distance $2$ units along the positive direction of the x-axis.

(c) Rotation through an angle $p/4$ about the origin in the anti clockwise direction.

The final position of the point is given by the co-ordinates

  1. $\left(\dfrac{4}{\sqrt{2}} , \dfrac{1}{\sqrt{2}}\right)$
  2. $\left(-\dfrac{1}{\sqrt 2} , \dfrac{7}{\sqrt 2}\right)$
  3. $\left(\dfrac{1}{\sqrt{2}} , \dfrac{7}{\sqrt{2}}\right)$
  4. $\left(-\dfrac{3}{\sqrt{2}} , \dfrac{4}{\sqrt{2}}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given Point $P(4,1)$


$(a)$ Reflection of given point about the line $y=x$
In Point P $x=4$ Hence about the line y-coordinate be $y=x\Rightarrow y=4$
Point P become $P(1,4)$

$(b)$ Transformation of Point P through distance $2$ units along the positive direction of the x-axis
$P(1+2,4)\equiv P(3,4)$

$(c)$Rotation of Point P through angle $\dfrac{\pi}{4}$ about origin in anti clockwise direction
$P\left (  r\cos\left ( \alpha+\dfrac{\pi}{4} \right ),r\sin\left ( \alpha+\dfrac{\pi}{4} \right )\right )$

Given point $P(3,4)$
$r=\sqrt{3^2+4^2}=5$
$\tan\alpha=\dfrac{4}{3}$
Hence In right angle triangle here hypotenuse be $5$
$\cos\alpha=\dfrac{3}{5}$ and $\sin\alpha=\dfrac{4}{5}$

$\cos\left ( \alpha+\dfrac{\pi}{4} \right )=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}$

$\cos\left ( \alpha+\dfrac{\pi}{4} \right )=\dfrac{3}{5}\times \dfrac{1}{\sqrt{2}}-\dfrac{4}{5}\times \dfrac{1}{\sqrt{2}}=-\dfrac{1}{5\sqrt{2}}$

$r\cos\left ( \alpha+\dfrac{\pi}{4}\right )=5\times \dfrac{-1}{5\sqrt{2}}=-\dfrac{1}{\sqrt{2}}$

$\sin\left ( \alpha+\dfrac{\pi}{4} \right )=\sin\alpha\cos\dfrac{\pi}{4}+\cos\alpha\sin\dfrac{\pi}{4}$

$\sin\left ( \alpha+\dfrac{\pi}{4} \right )=\dfrac{4}{5}\times \dfrac{1}{\sqrt{2}}+\dfrac{3}{5}\times \dfrac{1}{\sqrt{2}}=\dfrac{7}{5\sqrt{2}}$

$r\sin\left ( \alpha+\dfrac{\pi}{4}\right )=5\times \dfrac{7}{5\sqrt{2}}=\dfrac{7}{\sqrt{2}}$


Point be 

$P\left (  -\dfrac{1}{\sqrt{2}},\dfrac{7}{\sqrt{2}}\right )$


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

Image of the point $\left( -8,12 \right) $ with respect to the line mirror $4x+7y+13=0$ is 

  1. $\left( 16,2 \right) $
  2. $\left( -16,-2 \right) $
  3. $\left( -12,5 \right) $
  4. $\left( 12,-5 \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $(x _1,y _1)$ is $(-8,12)$


Let $(\alpha,\beta)$ be image of the point.


$a=4,\,b=7$ and $c=13$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2[4(-8)+7(12)+13]}{(4)^2+(7)^2}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2(-32+84+13)}{16+49}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2(65)}{65}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=-2$

$\Rightarrow$ $\dfrac{\alpha+8}{4}=-2$  and  $\dfrac{\beta-12}{7}=-2$

$\Rightarrow$  $\alpha+8=-8$   and  $\beta-12=-14$

$\Rightarrow$  $\alpha=-16$ and $\beta=-2$

So, the image of the point is $(-16,-2).$

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

If ${ P } _{ 1 }\left( \dfrac { 1 }{ 5 } ,\alpha  \right)$ and ${P } _{ 2 }\left( \beta ,\dfrac { 18 }{ 5 }  \right)$ be the images of point $P\left( 1,\gamma  \right)$ about lines ${ L } _{ 1 }:2x-y=\lambda$ and ${ L } _{ 2 }:2y+x=4$ respectively, then the value of $\alpha$is-

  1. $-\dfrac { 3 }{ 5 }$
  2. $\dfrac { 2 }{ 5 }$
  3. $\dfrac { 7 }{ 5 }$
  4. $-\dfrac { 8 }{ 5 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the reflection formula for a point across a line to find the coordinates of P1 and P2 in terms of lambda and gamma, then solve for alpha.