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 introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

  1. $x-2z=0$
  2. $2x-z=0$
  3. $2x+y=0$
  4. $x-2y=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the given points be A($1,2,3$) and B($3,2,-1$) and let the point equidistant from A and B be P($x,y,z$) 

then  $PA=PB$
$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}}  = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}} $
Squaring both sides
${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$
${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$
$-2x-6z+10=-6x+2z+10$
$4x-8z=0$
$x-2z=0$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The point equidistant from the points $(0,0,0), (1,0,0), (0,2,0)$ and $(0,0,3)$ is

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

Equation of sphere passes through origin is given by


$x^2+y^2+z^2+ax+by+cz=0 ...(1)$

Given, it also passes through $(1,0,0),(0,2,0),0,0,3)$

$1+a=0\Rightarrow a = -1$

$4+2b=0\Rightarrow b = -2$

and $ 9+3c=0\Rightarrow c = -3$

$\therefore$the equation of the sphere becomes $x^2+y^2+z^2-x-2y-3z=0$

Comparing with $x^2+y^2+z^2+2gx+2fy+2kz+C=0$

Therefore, the point equidistant from all the given four point will be the centre of the sphere $(1)$ passing through all these points which is $\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \  centre$.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b , a + b)$, then  $bx = ay$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a point (x,y) is equidistant from A(a,b) and B(c,d) then,

$x = \frac{{a + c}}{2},y = \frac{{b + d}}{2}$
So,
$\begin{array}{l}x = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{2},y = \frac{{\left( {b - a} \right) + \left( {a + b} \right)}}{2}\x = \frac{{2a}}{2},y = \frac{{2b}}{2}\x = a,y = b\end{array}$

putting values in bx=ay
Solving LHS,
b(a)=ab
Solving RHS,
a(b)=ab
As LHS=RHS the equation bx=ay is true.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Find the coordinates of the point on the $x$-axis that is equidistant from $P(4,3,1)$ and $Q(-2,-6,-2)$.

  1. $\displaystyle \left( \frac { 3 }{ 2 } ,0,0 \right) $
  2. $\displaystyle \left( -\frac { 3 }{ 2 } ,0,0 \right) $
  3. $\displaystyle \left( 0,-\frac { 3 }{ 2 } ,0 \right) $
  4. $\displaystyle \left( 0,\frac { 3 }{ 2 } ,0 \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $R(x,0,0)$ be the point on $x$-axis which is equidistant from $P(4,3,1)$ and $Q(-2,-6,-2)$

$\Rightarrow { \left( x-4 \right)  }^{ 2 }+{ \left( -3 \right)  }^{ 2 }+{ \left( -1 \right)  }^{ 2 }={ \left( x+2 \right)  }^{ 2 }+{ 6 }^{ 2 }+{ 2 }^{ 2 }$ gives $-12x=18$ 
So $x=-1.5$ 
Hence, $\displaystyle  R\equiv \left( -\frac { 3 }{ 2 } ,0,0 \right) $

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

A point $Q$ at a distance $3$ from the point $P(1,1,1)$ lying on the line joining the points $A(0,-1,3)$ and $P$, has the coordinates

  1. $(2,3,-1)$
  2. $(4,7,-5)$
  3. $(0,-1,3)$
  4. $(-2,-5,7)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Let the coordinates of point $Q$ be $(a,b,c)$


The distance of $Q$ from $P(1,1,1)$ is $3$


The equation of line $AP$ is $\displaystyle \frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{-2}$

Therefore the point on the line $AP$ will look like $Q(t+1,2t+1,1-2t)$

$|QP|=\sqrt{(t+1-1)^2+(2t+1-1)^2+(1-2t-1)^2}=\sqrt{9t^2}=3$

So $|QP| = \pm3t=3$

$\Rightarrow t=\pm1$

So the possible coordinates of $Q$ are $(2,3,-1)$ and $(0,-1,3)$

Therefore the correct options are $A$ and $C$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The points $(4, -5, 1)$, $(3, -4, 0)$, $(6, -7, 3)$, $(7, -8, 4)$ are vertices of a

  1. square

  2. parallelogram

  3. rectangle

  4. rhombus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $A=(4,-5,1)$
$B=(3,-4,0)$
$C=(6,-7,3)$
$D=(7,-8,4)$
Now let the quadrilateral be $ABCD$.
Then
$AB=-i+j-k$ $|AB|=\sqrt{3}$
$BC=3i-3j+3k$ $|BC|=3\sqrt{3}$
$CD=i-j+k$  $|CD|=\sqrt{3}$
$AD=3i-3j+3k$ $|AD|=3\sqrt{3}$.
Hence opposite sides are equal and parallel.
Therefore the above points form a parallelogram.
Now all the sides are not equal.
Hence it cannot qualify as a rhombus or square.
Now dot product of $AB$ and $BC$ is not zero.
Hence adjacent sides are not perpendicular to each other.
Therefore it is not a rectangle also.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

$A, B, C$ are three points on the axes of $x, y$ and $z$ respectively at distance $a, b, c$ from the origin $O$; then the co - ordinates of the point which is equidistant from $A, B, C$ and $O$ is

  1. $\displaystyle \left ( a,b,c \right )$
  2. $\displaystyle \left ( \frac{a}{2},\frac{b}{2},\frac{c}{2} \right )$
  3. $\displaystyle \left ( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right )$

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

Let $P$ be the required point $\displaystyle \left ( x,y,z \right )$ and the point
$A, B, C$ and $O$ are $\displaystyle \left ( a,0,0 \right ),\left ( 0,b,0 \right ),\left ( 0,0,c \right )$ and $\displaystyle \left ( 0,0,0 \right )$ ;

We are given that $\displaystyle PO=PA=PB=PC.$
Taking $\displaystyle PO=PA$ or $\displaystyle PO^{2}=PA^{2},$ we get
$\displaystyle x^{2}+y^{2}+z^{2}=\left ( x-a \right )^{2}+y^{2}+z^{2}$
$\displaystyle 0=a^{2}-2ax$  i.e.  $\displaystyle x=\dfrac {a}{2}$
Similarly taking $\displaystyle PO^{2}=PB^{2}$ and $\displaystyle PO^{2}=PC^{2},$ we get 
$\displaystyle y=\dfrac {b}{2}$ and $\displaystyle z=\dfrac {c}{2}$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

$P(0,5,6),Q(1,4,7),R(2,3,7)$ and $S(3,5,16)$ are four points in the space. The point nearest to the origin $O(0,0,0)$ is

  1. $P$
  2. $Q$
  3. $R$
  4. $S$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The $4$ points are as given.
We calculate their individual distance from the origin.
$OP =$ $ {({5}^{2} + {6}^{2})}^{0.5} $ = $ {(61)}^{0.5} $
$OQ =$ $ {({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5} $ = $ {(66)}^{0.5} $
$OR =$ $ {({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5} $ = $ {(62)}^{0.5} $
$OS =$ $ {({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5} $= $ {(290)}^{0.5} $
Hence, $P$ is the nearest to the origin.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

A point on the line $\displaystyle \frac{{x + 2}}{1} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 1}}{{2\sqrt 2 }}$ at a distance 6 from the point (2, 3, 1) is

  1. $(4-21, 1+12\sqrt{2})$
  2. $\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$
  3. $\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$
  4. $\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$
$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$
$\lambda^2+16\lambda^2+8\lambda^2=36$
$\lambda=\pm \frac {6}{5}$
$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The point equidistant from the point $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$ and $C(0, 0, c)$ has the coordinates

  1. $(a, b, c)$
  2. $(a/2, b/2, c/2)$
  3. $(a/3, b/3, c/3)$
  4. $(a/4, b/4, c/4)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$P(x, y, z)$
$PO=\sqrt{x^2+y^2+z^2}$
$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$
$PB=\sqrt{x^2+(b-y)^2+z^2}$
$PC=\sqrt{x^2+y^2+(C-z)^2}$
$\Rightarrow PO=PA$
$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$
$x^2+y^2+z^2=(a-x)^2+y^2+z^2$
$x^2=(a-x)^2$
$x=a-x$
$\Rightarrow x=\dfrac{a}{2}$
$y=b/2$
$z=c/2$.
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The locus of a point P which moves such that $PA^2-PB^2=2k^2$ where A and B are $(3, 4, 5)$ and $(-1, 3, -7)$ respectively is 

  1. $8x+2y+24z-9+2k^2=0$
  2. $8x+2y+24z-2k^2=0$
  3. $8x+2y+24z+9+2k^2=0$
  4. $8x-2y+24z-2k^2=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P{A}^{2}-P{B}^{2}=2{k}^{2}$
$\Rightarrow \left[{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}\right]-\left[{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}\right]=2{k}^{2}$
$\Rightarrow \left(x-3+x+1\right)\left(x-3-x-1\right)+\left(y-4+y-3\right)\left(y-4-y+3\right)+\left(z-5+z+7\right)\left(z-5-z-7\right)=2{k}^{2}$
$\Rightarrow \left(2x-2\right)\left(-4\right)+\left(2y-7\right)\left(-1\right)+\left(2z+1\right)\left(-12\right) =2{k}^{2}$
$\Rightarrow -8x+8-2y+7-24z-24=2{k}^{2}$
$\Rightarrow -8x-2y-24z-9=2{k}^{2}$
$\Rightarrow 8x+2y+24z+9+2{k}^{2}=0$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The points $A(1,2,-1),B(2,5,-2),C(4,4,-3)$ and $D(3,1,-2)$ are

  1. collinear

  2. vertices of a rectangle

  3. vertices of a square

  4. vertices of a rhombus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

AB${=}$ $\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$
AB${=}$ $\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$
AB${=}$ $\sqrt{11}$
Similarly you find that BC${=}$ $\sqrt{6}$  CD${=}$ $\sqrt{11}$  and DA${=}$$\sqrt{6}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$
AC${=}$ $\sqrt{17}$
similarly BD${=}$ $\sqrt{17}$ 
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through  BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The coordinates of a point which is equidistant from the point $(0,0,0),(a,0,0),(0,b,0)$ and $(0,0,c)$ are given by

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

Let $P(x,y,z)$ be the required point.

Then $OP=PA=PB=PC$
Now $OP=PA \Longrightarrow OP^2+PA^2 \Longrightarrow x^2+y^2+z^2$

= $(x-a)^2+(y-0)^2+(z-0)^2 \Longrightarrow x=\dfrac{a}{2}$
Similarily, $OP=PB=y=\dfrac{b}{2}$ and $OP=PC=z=\dfrac{c}{2}$

hence the coordinate of the required point are $\left(\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}\right)$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Find the locus of the point of intersection of the lines $\sqrt 3 x-y-4\sqrt 3\lambda=0$ and $\sqrt 3 \lambda x +\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

  1. $3x^2-y^2=48$
  2. $y^2-3x^2=24$
  3. $4x^2-3y^2=16$

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

Let $(h,k)$ be the point of intersection of the given lines. Then,

$\sqrt 3 h-k-4\sqrt 3 \lambda=0$ and $\sqrt3 \lambda h +\lambda k-4\sqrt 3=0$
$\sqrt 3 h-k=4\sqrt 3\lambda$ and $\lambda(\sqrt 3h+k)=4\sqrt 3$
$(\sqrt 3 h-k)\lambda (\sqrt 3h +k)=(4\sqrt 3\lambda)(4\sqrt 3)$
$3h^2-k^2=48$
Hence, the locus of $(h,k)$ is $3x^2-y^2=48$.