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 linear graphs quadrants the cartesian plane the cartesian system

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
Solve the equations of two lines to get a point and satisfy it in the third equation of a line.

$4ax+2ay+c=0$ and $5bx+2by+d=0$

Let the point of intersection be $(k, -k)$

From both equations, we get 

$4ak-2ak+c=0\ ,\ 5bk-2bk+d=0$

$2ak+c=0\ ,\ 3bk+d=0$

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

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

$\therefore 3bc-2ad=0$

Option C
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

Mid point of $A(0, 0)$ and $B(1024, 2050)$ is ${A _1}$. mid point of ${A _1}$ and B is ${A _2}$ and so on. Coordinates of ${A _{10}}$ are.

  1. $(1022, 2044)$
  2. $(1025, 2050)$
  3. $(1023, 2046)$
  4. $(1, 2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sequence of midpoints follows the rule A_n = (A_{n-1} + B) / 2. This is a geometric progression converging to B. After 10 steps, the coordinates are (1024 * (1 - (1/2)^10), 2050 * (1 - (1/2)^10)). Calculation: 1024 * (1023/1024) = 1023 and 2050 * (1023/1024) = 2046.009... which rounds to 2046.

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

If the coordinates of the extermities of diagonal of a square are $(2,-1)$ and $(6,2)$, then the coordinates of extremities of other diagonal are 

  1. $\left(\dfrac{5}{2},\dfrac{5}{2}\right)$
  2. $\left(\dfrac{11}{2},\dfrac{3}{2}\right)$
  3. $\left(\dfrac{11}{2},\dfrac{-3}{2}\right)$
  4. $\left(\dfrac{5}{2},-\dfrac{5}{2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Coordination\, \, of\, \, mid-po{ { int } }\, \, 0 \ =\left( { \frac { { 6+2 } }{ 2 } ,\frac { { 2-1 } }{ 2 }  } \right)  \ =\left( { 4,\frac { 1 }{ 2 }  } \right)  \ AB=BC \ \Rightarrow { \left( { { x _{ 1 } }-2 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }+1 } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-6 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-2 } \right) ^{ 2 } } \ \Rightarrow 8{ x _{ 1 } }+6{ y _{ 1 } }=35\to (i) \ AO=BO \ \Rightarrow { \left( { 2-4 } \right) ^{ 2 } }+{ \left( { -1-\frac { 1 }{ 2 }  } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-4 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-\frac { 1 }{ 2 }  } \right) ^{ 2 } } \ \Rightarrow 4+\frac { 9 }{ 4 } =x _{ _{ 1 } }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }+16+\frac { 1 }{ 4 }  \ \Rightarrow x _{ 1 }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }=-10\to (ii) \ from\, \, equation\, \, (i) \ put \ { x _{ 1 } }=\frac { { 35-6{ y _{ 1 } } } }{ 8 } \, \, in\, \, equation\, \, (ii) \ \Rightarrow { \left( { \frac { { 35-6{ y _{ 1 } } } }{ 8 }  } \right) ^{ 2 } }+y _{ 1 }^{ 2 }-\left( { 35-6{ y _{ 1 } } } \right) { y _{ 1 } }=-10 \ \Rightarrow 4y _{ 1 }^{ 2 }-4{ y _{ 1 } }-15=0 \ \Rightarrow \left( { 2{ y _{ 1 } }+3 } \right) \left( { { y _{ 1 } }-5 } \right) =0 \ \Rightarrow { y _{ 1 } }=\frac { { -3 } }{ 2 } ,5 \ { y _{ 1 } }=\frac { { -3 } }{ 2 } \to { x _{ 1 } }=\frac { { 35-6\times -\frac { 3 }{ 2 }  } }{ 8 } =\frac { { 11 } }{ 2 }  \ { y _{ 1 } }=5\to { x _{ 1 } }=\frac { { 35-6\times 5 } }{ 8 } =\frac { 5 }{ 8 }  \ The\, \, vertices\, of\, \, other\, \, two\, vertices\, \, are \ \left( { \frac { { 11 } }{ 2 } ,\frac { { -3 } }{ 2 }  } \right) \, \, and\, \, \left( { \frac { 5 }{ 8 } ,5 } \right)  \end{array}$

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

If sum of distance of a point from two perpendicular lines in a plane is $1$, then its locus is ?

  1. Square

  2. Circle

  3. A straight line

  4. An intersecting line
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x axis & y axis are the perpendicular lines. The sum of the distances from point $p(x, y)$ is $1$ 

i.e.,$|x| + |y| = 1$

The locus of the point 'p' which is the rhombus whose sides are $x + y = 1 ; -x + y = 1 ; x - y = 1 ; -x - y = 1$

$\bot r$ lines other than coordinate axis gives same result so locus is a square. 

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

The nearest point on the line $3x-4y=25$ from the origin is

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

Distance of the line $3x-4y-25=0$ from the origin is 
$\displaystyle d=\frac{|-25|}{\sqrt{25}}$
$\Rightarrow d=5$
Only the point given in option B lies on the given line .
Also, its distance from origin is 5.
So, (3,-4) is the nearest point on the line from the origin.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the projection of point P$(\vec{p})$ on the plane $\vec{r}\cdot \vec{n}=q$ is the points $S(\vec{s})$, then.

  1. $\vec{s}=\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$
  2. $\vec{s}=\vec{p}+\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$
  3. $\vec{s}=\vec{p}-\dfrac{(\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$
  4. $\vec{s}=\vec{p}-\dfrac{(\vec{q}-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the problem 

Let,
$\pi : \vec r.\vec n=q$ be the plane.
Now $S$ will lie on $\pi $. 
Consider $\vec S$ will joins $O(0,0,0)$ and  $S$. And as $P$ is projection on plane, 
then 
$\vec {PS}$ is perpendicular to plane.

$\vec P=\vec {OP}$
Now In triangle $OSP$, by triangle law of vector addition.

$\vec {OS}+\vec {SP}=\vec {OP}$  

$|\vec {SP}|=$ distance between $P$ and plane 
$=\dfrac{\vec p.\vec n-q}{|\vec n|}$

$\vec {SP}=|\vec {SP}|.\hat {SP}$

Now, 
$\hat {SP}=\hat n=\dfrac{\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{|\vec {SP}|\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{(\vec p.\vec n-q)\vec n}{|\vec n|^2}$

$\vec {OS}+\vec {OP}=\vec {OP}$
$\vec S+\vec {SP}=\vec P$

$\vec S=\vec p-\vec {SP}$
$=\vec p-(\dfrac{\vec p.\vec n-q}{|\vec n|^2})\vec n$

$\vec S=\vec p+\dfrac{(q-\vec p.\vec n)\vec n}{|\vec n|^2}$

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

The coordinates of any point, which lies on $x$ axis are

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

In 3-dimensional plane, the point which lies on $x$-axis does not have any part in y and z axes.

At that point, the value of $y$ and $z$ will be $0$.
Hence, coordinate of any point which lies on $x$ axis are $(x,0,0)$.

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

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$