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 maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

If point $p$ lies in first octant, then the sign of $x-$ coordinate will always be 

  1. $+$
  2. $-$
  3. $x$ coordinate is always $0$
  4. $x$ coordinate can be $+$ or $-$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the first octant, the values of x,y and z axes are positive.

Any point which lies in first octant has all their coordinate values as positive.
Since point $p$ lies in first octant, so, p will have all its coordinates as positive.
Hence, sign of $x$-coordinate will always be $+$.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The coordinate of any point, which lies in $xy$ plane , is

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

Given that the point lies in $xy$ plane

In $xy$ plane , the coordinate of $z$ will be zero
So $(x,x,0)$ represents a point which lies in $xy$ plane
Therefore option $B$ is correct

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The points $(3,\ 2,\ 0),\ (5,\ 3,\ 2)$ and $(-9,\ 6,\ -3)$, are the vertices of a triangle $ABC.AD$ is the internal bisector of $\angle\ BAC$ which meets $BC$ at $D$. Then the co-ordinates of $D$, are

  1. $\left[ {\dfrac{{17}}{{16}},\ \dfrac{{57}}{{16}},\ \dfrac{{19}}{8}} \right]$
  2. $\left[ {\dfrac{{19}}{{8}},\ \dfrac{{57}}{{16}},\ \dfrac{{17}}{16}} \right]$
  3. $\left[0,\ 0,\ {\dfrac{{17}}{{16}}}\right]$
  4. $\left[{\dfrac{{17}}{{16}}},\ 0,\ 0\right]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$AD$ is bisector of $ \angle BAC $ 

$ \Rightarrow $ Ratio at $D$ is $ c:b$ where 

$ c = AB = \sqrt{(3-5)^{2}+(2-3)+(0-2)^{2}} $

$ = \sqrt{4+1+4} = \sqrt{9} = 3 $

$ b = AC = \sqrt{(3+9)^{2}+(2-6)^{2}+(0+3)^{2}} $

$ = \sqrt{144+16+9} = \sqrt{162} = 13 $

for point $D$ 

$ x = \dfrac{c(-9)+b(6)}{c+b} = \dfrac{3(-9)+13(5)}{3+13} = \dfrac{38}{16} = \dfrac{19}{8} $

$ y = \dfrac{c(6)+b(3)}{c+b} = \dfrac{3(6)+13(3)}{3+13} = \dfrac{57}{16} $

$ z = \dfrac{c(-3)+b(2)}{c+b} = \dfrac{3(-3)+13(2)}{3+13} = \dfrac{17}{16} $

Hence, point $D$ is $ [\dfrac{19}{8},\dfrac{57}{16},\dfrac{17}{6}] $ 
Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The points (-5,12), (-2,-3),(9,-10),(6,5) taken in order, form

  1. Parallelogram

  2. rectangle

  3. rhombus

  4. square
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given points $A(-5, 12)\quad B(-2, -3), C(9, -10), D(6, 5)$


Distance $AB=\sqrt{(-5+2)^2+(12+3)^2}=\sqrt{9+225}=\sqrt{234}$
Distance $BC=\sqrt{(-2-9)^2+(-3+10)^2}=\sqrt{121+49}=\sqrt{170}$

Distance $CD=\sqrt{(9-6)^2+(-10-5)^2}=\sqrt{9+225}=\sqrt{234}$
Distance $AD=\sqrt{(-5-6)^2+(12-5)^2}=\sqrt{121+49}=\sqrt{170}$

Distance $AC=\sqrt{(-5-9)^2+(12+10)^2}=\sqrt{196+484}=\sqrt{680}$
Distance $BD=\sqrt{(-2-6)^2+(-3-5)^2}=\sqrt{64+64}=\sqrt{128}$

These points forms a parallelogram, opposite pair of sides are equal and adjacent sides do not form right angles.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

If G is the centroid of $\triangle ABC$ and BC = 3, CA = 4, AB = 5 then BG =

  1. $\dfrac { \sqrt { 73 } }{ 3 } $
  2. $\dfrac { \sqrt { 13 } }{ 3 } $
  3. $\dfrac { \sqrt { 52 } }{ 3 } $
  4. $\dfrac { \sqrt { 26 } }{ 3 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The triangle with sides 3, 4, 5 is a right triangle. Using Apollonius theorem or coordinate geometry, the length of the median to side AB (c=5) is m_c = 1/2 * sqrt(2a^2 + 2b^2 - c^2) = 1/2 * sqrt(2*16 + 2*9 - 25) = 1/2 * sqrt(7) = sqrt(7)/2. The centroid G divides the median in a 2:1 ratio, so BG = 2/3 * m_c = 2/3 * sqrt(7)/2 = sqrt(7)/3. However, checking the options, sqrt(52)/3 is 2*sqrt(13)/3. Let's re-verify: median to AB is 1/2 * sqrt(2*16 + 2*9 - 25) = sqrt(7)/2. The distance BG is 2/3 of the median. None match perfectly. Re-evaluating: maybe median to BC? m_a = 1/2 * sqrt(2*16 + 2*25 - 9) = 1/2 * sqrt(32+50-9) = sqrt(73)/2. BG = 2/3 * sqrt(73)/2 = sqrt(73)/3. Option A is sqrt(73)/3. Wait, the question asks for BG, which is the segment from vertex B to centroid G. This is 2/3 of the median from B to AC. Median m_b = 1/2 * sqrt(2*a^2 + 2*c^2 - b^2) = 1/2 * sqrt(2*9 + 2*25 - 16) = 1/2 * sqrt(18+50-16) = 1/2 * sqrt(52) = sqrt(52)/2. BG = 2/3 * sqrt(52)/2 = sqrt(52)/3.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

Arrange the points: $\mathrm{A}(1,2-3), \mathrm{B}(-1,2,-3), \mathrm{C}(-1,-2-3)$ and $\mathrm{D}(1,-2, -3)$ in the increasing order of their octant numbers:

  1. $A,B,C,D$
  2. $B,C,D,A$
  3. $C,D,A,B$
  4. $D,C,B,A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 Octant $I$  $II$ $III$  $IV$  $V$  $VI$  $VII$ $VIII$
 Signs: $+,+,+$  $-,+,+$ $-,-,+$  $+,-,+$  $+,+,-$ $-,+,-$ $-,-,-$  $+,-,-$ 

Based on this, increasing order is

$ A,B ,C,D$

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The point $(3, 0, -4)$ lies on the

  1. Y-axis

  2. Z-axis

  3. XY-plane

  4. XZ-plane

  5. YZ-plane
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$(3, 0, -4)$   $\rightarrow$   Given point
Clearly, $y = 0$ and $ x$ and $z$ have non-zero value.
If the point lies on $x-z$ plane, this condition is possible.
Hence, the answer is $XZ$- plane.
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

A point on the line $\bar {r}=2\hat {i}+3\hat {j}+4\hat {k}+t(\hat {i}+\hat {j}+\hat {k})$ is

  1. $(2014,2015,2016)$
  2. $(2013,2015,2017)$
  3. $(2013,2014,2017)$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A point on the line is given by (2+t, 3+t, 4+t). For option A, 2+t=2014 => t=2012. Then 3+2012=2015 and 4+2012=2016. This matches (2014, 2015, 2016).

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The position vector of point $A$ is $(4, 2, -3)$. If $p {1}$ is perpendicular distance of $A$ from $XY-plane$ and $p _{2}$ is perpendicular distance from Y-axis, then $p _{1} + p _{2} =$ ______.

  1. $8$
  2. $3$
  3. $2$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P _1$ is perpendicular distance of A from XY-plane 
So, $x _1=0,y _1=0,z _=-3$
$\therefore p _1=\sqrt{(-3)^2}=3$
$p _2$ is perpendicular distance from Y-axis
So, $x _1=4,y _1=0,z _1=-3$
$p _{2} = \sqrt {x _{1}^{2} + y _{1}^{2}}$
$\therefore p _2=\sqrt{4^2+(-3^2)}=5$
$p _{1} + p _{2} = 5 + 3 = 8$.
Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The co-ordinate of a point where the line $(2, -3, 1)$ and $(3, -4, -5)$ cuts the plane $2x + y + z = 7$ are $(1, k, 7)$ then value of $k$ equals

  1. $1$
  2. $-2$
  3. $2$

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

We know that the equation of the line passing through the point $(x _1,y _1,z _1) , (x _2,y _2,z _2) $


$ \dfrac{x-x _1}{x _2-x _1} = \dfrac {y-y _1}{y _2-y _1} =  \dfrac {z-z _1}{z _2-z _1}$

Now the given line passes through the points $(2,−3,1) , (3,−4,−5)$


Hence the equation of the line is $ \dfrac{x-2}{3-2} = \dfrac {y+3}{-4+3} =  \dfrac {z-1}{-5-1}$
`
= $ \dfrac{x-2}{1} = \dfrac {y+3}{-1} =  \dfrac {z-1}{-6}$

Let the above equation be equal to :

$ \dfrac{x-2}{1} = 1$ .... (1)

$\dfrac {y+3}{-1} = a$ .....(2)

$\dfrac {z-1}{-6} = 7$ .....(3)

$\Rightarrow$ from eqn (2) 

$y=−4+a$

cube cuts the plane $2x+y+z=7$ ... (4)

substitute the $(1, k ,7)$ in eqn (4)

$2(1) + (-4+a) + 7 = 7$

$a = 2$

substitute a value in $y=−4+a$
we get $y = -4+2 = -2$

hence the ans will be $(1 , -2 , 7)$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

 Consider a point $P (1, 2, 3)$, plane $ \pi : x + y + z = 11 $ and the line $ L : \displaystyle \frac{x+1}{1}=\displaystyle \frac{y-12}{-2} = \displaystyle \frac{z-7}{2} $ The foot of the $ \perp  $ drawn from the point P meet the plane $ \pi $ at M, then co-ordinate of M is

  1. $ \left ( \displaystyle \frac{8}{3},\:\displaystyle \frac{-11}{3},\:\displaystyle \frac{14}{3} \right ) $
  2. $ \left ( \displaystyle \frac{-8}{3},\:\displaystyle \frac{-11}{3},\:\displaystyle \frac{-14}{3} \right ) $
  3. $ \left ( \displaystyle \frac{8}{3},\:\displaystyle \frac{11}{3},\:\displaystyle \frac{14}{3} \right ) $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P = (1,2,3) \rightarrow$ Point
$\Rightarrow\pi  : x+y+z = 11$
     $L : \dfrac { x+1 }{ 1 } =\dfrac { y-12 }{ -2 } =\dfrac { z-7 }{ 2 }$
Foot of perpendicular to the plane :
$\Rightarrow\dfrac { h-{ x } _{ 1 } }{ a } =\dfrac { k-{ y } _{ 1 } }{ b } =\dfrac { l-{ z } _{ 1 } }{ c } =\dfrac { -(a{ x } _{ 1 }+b{ y } _{ 1 }+c{ z } _{ 1 }) }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }$
     $\dfrac { h-1 }{ 1 } =\dfrac { k-2 }{ 1 } =\dfrac { l-3 }{ 1 } =\dfrac { 1+2+3 }{ 3 }$
     $h-1 = k-2= l-3= -2$
     $h= -1, k=0, l=1$
The required point is
$(h+\dfrac { 11 }{ 3 } , k+\dfrac { 11 }{ 3 } , l+\dfrac { 11 }{ 3 } )$
$(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } )$
Hence the answer is $(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } ).$