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 direction cosines and direction ratios three dimensional geometry maths

The values of $a$ for which point $(8, -7, a), (5, 2, 4)$ and $(6, -1, 2)$ are collinear.

  1. $-4$
  2. $-2$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the points be $A(8,-7,a), B(5,2,4), C(6,-1,2)$
Direction ratio of $BC =(1,-3,-2)$
If $A,B,C$ are collinear points then direction ratio of $BC$ and $AB$ must be proportional.
Direction ratio of $AB=(-3,9.4-a)$
$\therefore -3=\lambda1$
$\therefore 9=\lambda(-3)$
$\therefore 4-a=\lambda(-2)$ and $\lambda=-3$
$\therefore \dfrac{4-a}{-3}=-2$

$\therefore a=-2$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The point collinear with $(4, 2, 0)$ and $(6, 4, 6)$ among the following is

  1. $(0,4,6)$
  2. $(8,6,8)$
  3. $(1, -4, -6)$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Cartesian form  of a line passing through $(x _1,y _1,z _1)$  and  $(x _2,y _2,z _2) $ is:
$\dfrac{x-x _1}{x _2-x _1}=\dfrac{y-y _1}{y _2-y _1}=\dfrac{z-z _1}{z _2-z _1}=\lambda$

Cartesian form  of the line passing through (4,2,0) and (6,4,6) is:
$\dfrac{x-4}{6-4}=\dfrac{y-2}{4-2}=\dfrac{z-0}{6-0}=\lambda$

$\dfrac{x-4}{2}=\dfrac{y-2}{2}=\dfrac{z-0}{6}=\lambda$

Therefore general point on the line is $(2\lambda+4,2\lambda+2,6\lambda)$
Now if we compare this coordinate with given options,we get different values of $\lambda$.

that means none of the points given in options lie on the line(Non-Colinear).

Therefore, (D) option is correct.
Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(0, 1, -2), (3$, $\lambda$,$ 1)$ and ($\mu$, $7, 4$) are collinear, the point on the same line is

  1. $(5, 6, 3)$
  2. $(1, -1, -2)$
  3. $(-5, -6, -3)$
  4. $(0, 0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The direction vector using the first two points we get $3i+(\lambda-1)j+3k$. 
Using the first and third point $\mu i+6j+6k$. 
Since they are up to multiplication by a constant we get $\lambda=4$ and $\mu=6$.
 Hence direction vector is $i+j+k$. 
Any point on the line is given by $(0,1,-2)+t(1,1,1)$. Plugging $t=5$ we get the point $(5,6,3)$.   

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Given $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$

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

Given:- $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ 

The equation of line $AB$ is given by $r=i-j+t(2i+2j+2k)$ or $\dfrac{x-1}{2} = \dfrac{y+1}{2} = \dfrac{z}{2} $ 
The points $(2,0,1)$ and $(0,-2,1)$ lies on $AB$. 
The equation of line $CD$ is given by $r=2i-2j+4k+t(3i-3j+5k)$ or $ \dfrac{x-2}{3} = \dfrac{y+2}{-3} = \dfrac{z-4}{5}$
$(2,-2,4)$ lie on $CD$. 
$(2,2,4)$ does not lie on any of the lines AB or CD.
Hence, option A is correct.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $a(1, 2, -1), B(2, 6, 2)$ and $c(\lambda, -2, -4)$ are collinear then $\lambda$ is

  1. $0$
  2. $2$
  3. $-2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

D.R of AB are $ 2 -1, 6-2,2-(-1) i.e. 1,4,3$

D.R. of AC are $ λ.−1,−2−2,−4−(−1) $
$i.e., λ−1,−4,−3 $
Since A, B, C are collinear $\therefore AB||BC $
$\therefore \dfrac {\lambda-1}{1} =\dfrac{-4}{4}=\dfrac{-3}{3}$
$\Longrightarrow \lambda-1=-1$
$\therefore \lambda=0$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points (p. 0), (0, q) and (1, 1) are collinear then $\dfrac { 1 }{ p } +\dfrac { 1 }{ q } $ is equal to 

  1. -1

  2. 1

  3. 2

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are $(p, o)(o, q)(i,q)$

$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$

$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$
$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$

$\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

Given $A(1,-1,0)$; $B(3,1,2)$; $C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$?

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

$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\ \vec { AB } = <3-1, 1-(-1), 2-0>$ 

       $= <2,2,2>$ 
$\therefore \quad Equation\quad of\quad line\quad AB:\ \dfrac { x-1 }{ 2 } =\dfrac { y-(-1) }{ 2 } =\dfrac { z-0 }{ 2 } \quad \Longrightarrow \quad \dfrac { x-1 }{ 2 } =\dfrac { y+1 }{ 2 } =\dfrac { z-0 }{ 2 }$ 
$\vec { CD } = <-1-2, 1-(-2), -1-4>$ 
        $= <-3,3,-5>$ 
$\therefore \quad Equation\quad of\quad line\quad CD:\ \dfrac { x-2 }{ -3 } =\dfrac { y-(-2) }{ 3 } =\dfrac { z-4 }{ -5 } \quad \Longrightarrow \quad \dfrac { x-2 }{ -3 } =\dfrac { y+2 }{ 3 } =\dfrac { z-4 }{ -5 }$ 
By putting the values of points given in the choices in the equation of lines $AB$ and $CD$, $(2,2,4)$ is the point which neither lie on $AB$ nor on $CD$.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $A(1,2,-1)$, $B(2,6,2)$ and $\displaystyle C\left ( \lambda,-2,-4 \right )$ are collinear, then $\displaystyle \lambda $ is

  1. $0$
  2. $2$
  3. $-2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

D.R. of AB are $2−1,6−2,2−(−1)i.e.,1,4,3.$

D.R. of AC are $λ.−1,−2−2,−4−(−1) $
$i.e., λ−1,−4,−3 $
Since A, B, C are colinear, 
$ \therefore AB||BC$
$\therefore \dfrac{λ−1}{1}=\dfrac{−4}{4}=\dfrac{−3}{3}⇒λ−1=−1⇒λ=0$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the three points with position vectors $\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c} $ are collinear, then $\displaystyle \lambda= $

  1. <font color="#888888">$1$</font>

  2. <span class="MathJax_Preview"><span class="MJXp-math"><span class="MJXp-mn">2
  3. $3$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given vectors are collinear, so $l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$
Comparing the coefficients of $\bar{a} \rightarrow l + 2k = 0 $
$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$
$\bar{c} \rightarrow 3l - 4k = 10l + 10k$
$\Rightarrow l = -2k$ and so $\lambda = 3$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

For what value of $m$, the points $(3,5)$, $(m,6)$ and $\begin{pmatrix} \dfrac { 1 }{ 2 },\dfrac {15 }{ 2 } \end{pmatrix}$ are collinear?

  1. $9$
  2. $5$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. 
Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ $=$ $ $\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $
So, slope of the line joining $ (3,5) , (m,6) = $ Slope of the line joining $ (3,5) $ and $\left  (\dfrac {1}{2}, \dfrac {15}{2}\right ) $ 

Therefore, $ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $
$\Rightarrow  \dfrac { 1 }{ m - 3 } = -1 $

$\Rightarrow  m - 3 = -1 $

$\Rightarrow  m = 2 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(p,0)$, $(0,q)$ and $(1,1)$ are collinear, then $\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$ is equal to:

  1. $-1$
  2. $1$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 }

\right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ = $ $\dfrac { { y

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

So, slope of the line joining $ (p,0) , (0,q) = $ Slope of the line joining

$ (0,q) $ and $ (1,1) $

$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $

$ - \dfrac { q }{ p } = 1 - q $

Dividing both sides by $q$,
$ - \dfrac { 1 }{ p } =  \dfrac { 1 }{ q } - 1 $

$ => \dfrac { 1 }{ p } +  \dfrac { 1 }{ q } = 1 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Determine if the points $(1,5)$ $(2,3)$ and $(-2,-11)$ are collinear.

  1. True

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

The given points are $A(1,5)$, $B(2,3)$ and $C(-2,-11)$.


Let us calculate the distance : $AB$, $BC$ and $CA$ by using distance formula.

$AB =\sqrt { (2-1)^{ 2 }+(3-5)^{ 2 } } =\sqrt { (1)^{ 2 }+(-2)^{ 2 } } $

$=\sqrt {1+4} = \sqrt{ 5 }$ units

$BC =\sqrt { (-2-2)^{ 2 }+(-11-3)^{ 2 } }=\sqrt { (-4)^{ 2 }+(-14)^{ 2 } }$

$=\sqrt {16+196} =\sqrt {212} = 2\sqrt{53}$ units

$CA =\sqrt { (-2-1)^{ 2 }+(-11-5)^{ 2 } }$

$=\sqrt { (-3)^{ 2 }+(-16)^{ 2 } } =\sqrt {9+256} = \sqrt {265 }$ 

$=\sqrt {5}\times\sqrt {53}$ units

From the above we see that : $AB+BC\neq CA$

Hence, the above stated points $A(1,5)$, $B(2,3)$ and $C(-2,-11)$ are not collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

In each of the following find the value of $k$, for which the points are collinear.
(i) $(7,-2)$, $(5,1)$, $(3,k)$
(ii) $(8,1)$, $(k,-4)$, $(2,-5)$

  1. (i) $k = 4$
  2. (i) $k = 5$
  3. (ii) $k = 3$
  4. (ii) $k = 2$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x } _{ 1 },{ y } _{ 1 })$ ; $({ x } _{ 2 },{ y

} _{ 2 })$  and $({ x } _{ 3 },{ y } _{ 3 })$  is $ \left| \dfrac { {

x } _{ 1 }({ y } _{ 2 }-{ y } _{ 3 })+{ x } _{ 2 }({ y } _{ 3 }-{ y } _{ 1 })+{ x } _{

3 }({ y } _{ 1 }-{ y } _{ 2 }) }{ 2 }  \right| $


1) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (7,-2) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (5,1) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (3,k)$

In the area formula, we get

$ \left| \dfrac { 7(1-k) + 5(k+2) + 3(-2-1) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 7 -7k + 5k + 10 - 9 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -2k }{ 2 }  \right|  =

0 $

$ \Rightarrow  8 - 2k = 0 $

$ \Rightarrow  k = 4 $

2) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (8,1) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (k,-4) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (2,-5)$ in the area formula, we get


$ \left| \dfrac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -6k +10 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 18 -6k }{ 2 }  \right|  =

0 $

$ \Rightarrow  18 - 6k = 0 $

$ \Rightarrow  k = 3 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Are the points (1, 1), (2, 3) and (8, 11) collinear ?

  1. collinear

  2. Non collinear

  3. coplaner

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

Area of triangle formed by these vertices is 
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 2 & 3 & 1 \ 8 & 11 & 1 \end{vmatrix}$
Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 0 \ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$
Hence points are non collinear 

Multiple choice direction cosines and direction ratios three dimensional geometry maths

A point $P$ lies on a line whose ends are $A(1,2,3)$ and $B(2,10,1).$ If $z$ component of $P$ is $7,$ then the coordinates of $P$ are

  1. $(-1,-14,7)$
  2. $(1,-14,7)$
  3. $(-1,14,7)$
  4. $(1,14,7)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of line passing through $A$ and $B$ is $\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$

Substitute $z=7$, we get
$\Rightarrow \displaystyle \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { 7-3 }{ -2 } =-2$
Solve it to get $x=1-2=-1$, $y=2-16=-14$
So $P:$ $\left( -1,-14,7 \right) $