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

There are three points with position vectors $ -2a+3b+5c, a+2b+3c $ and$ 7a-c$. What is the relation between the three points?

  1. Collinear

  2. Forms a triangle

  3. In different plane

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

The relation between the three points are collinear

Thus option A is correct answer 

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 mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^{2}-4xy-2y^{2}-3x-2y-1=0$ is 

  1. $\left(\dfrac{5}{2},3\right)$
  2. $(-4,\dfrac{3}{2})$
  3. $ (-2,3)$
  4. $ (2,3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation is $3x^{2}-4xy-2y^{2}-3x-2y-1=0$

Let $\left({x} _{1},{y} _{1}\right)$ be a point to which the origin is shifted by translation

Let $\left(X,Y\right)$ be the new coordinates of the point $\left(x,y\right)$

$\therefore\,$ the equations of the transformation are $x=X+{x} _{1},\,y=Y+{y} _{1}$

Now the transformed equation is 
$3{\left(X+{x} _{1}\right)}^{2}-4\left(X+{x} _{1}\right)\left(Y+{y} _{1}\right)-2{\left(Y+{y} _{1}\right)}^{2}-3\left(X+{x} _{1}\right)-2\left(Y+{y} _{1}\right)-1=0$

$\Rightarrow\,3\left({X}^{2}+2X{x} _{1}+{{x} _{1}}^{2}\right)-4\left(XY+X{y} _{1}+Y{x} _{1}+{x} _{1}{y} _{1}\right)-2\left({Y}^{2}+{{y} _{1}}^{2}+2Y{Y} _{1}\right)-3X-{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,3{X}^{2}+3{{x} _{1}}^{2}+6X{x} _{1}-4X{y} _{1}-4{x} _{1}Y-4{x} _{1}{y} _{1}-2{Y}^{2}-2{{y} _{1}}^{2}+4Y{y} _{1}-3X-3{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,\left(3{X}^{2}-4XY-2{Y}^{2}\right)+\left(3{{x} _{1}}^{2}-2{{y} _{1}}^{2}-4{x} _{1}{y} _{1}-3{x} _{1}-2{y} _{1}-1\right)+2X\left(3{x} _{1}-2{y} _{1}-\dfrac{3}{2}\right)+2Y\left(-2{x} _{1}+2{y} _{1}-1\right)=0$

Solving the first degree terms,we have
$3{x} _{1}-2{y} _{1}=\dfrac{3}{2}$

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

Adding the above equations, we get
$3{x} _{1}-2{y} _{1}-2{x} _{1}+2{y} _{1}=\dfrac{3}{2}+1$

$\Rightarrow\,{x} _{1}=\dfrac{5}{2}$

From equation ,$-2{x} _{1}+2{y} _{1}=1$

$\Rightarrow\,2{y} _{1}=1+2{x} _{1}=1+2\times\dfrac{5}{2}=1+5=6$

$\Rightarrow\,{y} _{1}=\dfrac{6}{2}=3$

$\therefore\,\left({x} _{1},{y} _{1}\right)=\left(\dfrac{5}{2},3\right)$


Hence the point is $\left(\dfrac{5}{2},3\right)$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(-5, 0)$ is $2$ is:

  1. $\dfrac{x^2}{1}+\dfrac{y^2}{24}=1$
  2. $\dfrac{x^2}{24}+\dfrac{y^2}{1}=1$
  3. $\dfrac{x^2}{24}-\dfrac{y^2}{2}=1$
  4. $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The locus is nothing but hyperbola.
Difference of distance of a point from foci $=2a$ 

Given distance is $2 \Rightarrow a=1$
Distance between foci $=2ae=2\sqrt{a^2+b^2}=\sqrt{(5+5)^2}$
                                                 $\Rightarrow a^2+b^2 =25$
                                                  $\Rightarrow b^2=24$
Therefore, locus is $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$ 

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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. $4x^2-y^2=48$
  2. $x^2-4y^2=48$
  3. $3x^2-y^2=48$
  4. $y^2-3x^2=48$
Reveal answer Fill a bubble to check yourself
C 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 $\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$


$\sqrt{3}h-k=4\sqrt{3}\lambda$ and $\lambda(\sqrt{3}h+k)=4\sqrt{3}$

$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$

$3h^2-k^2=48$

Hence, the locus of (h,k) is $3x^2-y^2=48$.

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 $(\dfrac { 8 }{ 3 } , \dfrac { 11 }{ 3 } ,\dfrac { 14 }{ 3 } ).$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The locus of the points which are equidistant from $(-a, 0)$ and $x=a$ is

  1. $y^2=4ax$
  2. $y^2+4ax=0$
  3. $x^2+4ay=0$
  4. $x^2-4ay=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $(h,k)$ be any point on the curve.
Distance of this point from $(-a,0)=\sqrt{(h-(-a))^2+(k-0)^2}$
Distance of point $(h,k)$ from the line $x-a=0$ is $\dfrac{h-a}{1}=h-a$
Point $(h,k)$ is equidistant from $(-a,0)$ and the line $x-a=0$
$\implies  \sqrt{(h-(-a))^2+(k-0)^2}=h-a$

$\implies \sqrt{(h+a)^2+(k-0)^2}=h-a$

Squaring the above equation, we get
$\implies (h+a)^2+(k-0)^2=(h-a)^2$
$\implies h^2+a^2+2ah+k^2=h^2+a^2-2ah$

$\implies k^2=-4ah$
Subtitute $k=y$ and $h=x$, we get

$\implies y^2=-4ax$
$\implies y^2+4ax=0$
So, the answer is option (B)


Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

If a point $\mathrm{P}$ moves such that the distance from the point $\mathrm{A} (1, 1)$ and the line $x+y+2=0$ are equal then the locus of $\mathrm{P}$ is equal to

  1. a straightline

  2. a parabola

  3. a pair of st. lines

  4. an ellipse
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the coordinate at $P$ is $(h, k)$

$\therefore \ $ Distance com point $A(1, 1)$ is $AP=\sqrt {(n-1)^2+(k-1)^2}$

$\therefore \ $ Distance com line $x+y+2=0$ is $=\dfrac {h+k+2}{\sqrt {1^2 +1^2}}$

According to equation

$\sqrt {(h-1)^2 +(k-1)^2}=\dfrac {h+k+2}{\sqrt 2}$

Squaring both are

$2\left\{(h-1)^2 +(k-1)^2\right\}=(h+k+2)^2$

$\Rightarrow \ 2(h^2-2h+1+k^2-2k+1)=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ 2h^2-4h+2k^2-4k+4=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ h^2-hk+k^2=8h+4k$

$\Rightarrow \ (h-k)^2=8(h+k)$

Locus of $P$ is, $(x-y)^2=8(x+y)$ which is a parabola

Option $\to (B)$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes, then:

  1. $2bc-3ad =0$
  2. $2bc+3ad =0$
  3. $3bc -2ad =0$
  4. $3bc +2ad =0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If it lies in the fourth quadrant, we get$(x,-x)$

$2ax+c = 0$ and $3bx+d = 0$
$\cfrac{c}{2a} = \cfrac{d}{3b}$
$3bc-2ad = 0$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $a,b,c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes then

  1. $2bc-3ad=0$
  2. $2bc+3ad=0$
  3. $3bc-2ad=0$
  4. $3bc+2ad=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Since point of intersection lies in the fourth quadrant and is equidistant from coordinate axes,

the $x$ and $y$ co-ordinates will be same 

Hence the coordinates become $(h,-h)$

Passing $4ax+2ay+c=0$ through $(h,-h)$ 

$\Rightarrow 4ah-2ah+c=0$

$\Rightarrow h=-\dfrac{c}{2a}----------(1)$

Also passing the second line $5bx+2by+d=0$ through $(h,-h)$

$\Rightarrow 5bh-2bh+d=0$

$\Rightarrow h=-\dfrac{d}{3b}----(2)$

From eq (1) and (2)

$-\dfrac{c}{2a}=-\dfrac{d}{3b}$

$\Rightarrow 3bc-2ad=0$