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 mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The points of intersection of asymptotes with directrices lies on

  1. Auxillary circle

  2. Director circle

  3. Transverse axis

  4. Conjugate axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  • Asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.
  • A perpendicular drawn from the foci on either asymptote meet it in the same points as the corresponding directrix and the common points of intersection lie on the auxiliary circle.
Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

Point $A(2,1),B(3,-7),C$ is any point on the line $3x-2y=1$, then locus of point $D$ such that$ABCD$ is a parallelogram

  1. $3x-2y=20$
  2. $3x-y=20$
  3. $2x+3y=20$
  4. $3x-2y+18=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a parallelogram ABCD, the diagonals bisect each other. The midpoint of AC must equal the midpoint of BD. Using this property with C on the line 3x-2y=1, we derive the locus of D.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $60\hat{i}+3\hat{j}$, $40\hat{i}-8\hat{j}$, $a\hat{i}-52\hat{j}$  are collinear if

  1. $a=-40$
  2. $a=40$
  3. $a=20$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

suppose ${60i + 3j}$ , ${40i - 8j}$ and ${ai - 52j}$ is the three position of vector $A,B,C$


$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \ \overrightarrow { AB } =-20i-11j \ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \ \left( { a-40 } \right) i-44j=-20im-11jm \ -44=-11m \ m=\frac { { -44 } }{ { -11 } }  \ m=4 \ a-40=-20m \ a-40=-20\left( 4 \right)  \ a=-80+40 \ a=-40 \end{array}$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points $i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$ are

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

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

$\begin{matrix} A& B& C& D\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$
$\overline{AC} = (2-1)i + (2-1)j + k-k$
$=i+j$
$\overline{AB} = o + j - \overline{k} = j - \overline{k}$
$\overline{AD} = i + 2j + k$
$\begin{vmatrix} 1&1&0 \0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$
$=3-1 = 2 \neq 0$
Non coplanar.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $ 60i + 3j,  40i -8j$ and $ ai -52j $ are collinear if

  1. $a = -40$
  2. $a = 40$
  3. $a = 20$

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

Denoting $a,b,c$ by the given vectors respectively
These vectors will be collinear if there is some constant $k$ such that $c-a=K\left( b-a \right) $
$\Rightarrow a-60=-20K$ and $-55=-11K$
$\Rightarrow a=-100+60=-40$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The three points $ABC$ have position vectors $(1,x,3),(3,4,7)$ and $(y,-2,-5)$ are collinear then $(x,y)=$

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

$(1,x,3)=\lambda(3,4,7) + \mu (y,-2,-5)$
$1=3\lambda +\mu y$
$x= 4\lambda +(-2\mu)$
$3 = 7\lambda -5 \mu$
$2-x= (-3-y)\mu$
So only $x=2$,  $y=-3$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the three points  $A(\overline a),B(\overline  b),C(\overline c) $ are collinear ,the line passing through them is

$\overline r=\overline a+\lambda(\overline b-\overline a)$ then value of $\lambda $ is 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given line
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$\vec{r}=(1-\lambda)\vec{a}+\lambda\vec{b}$
if $a$ and $b$ are collinear then 
$xa+yb=0$
$x=1-\lambda$
$y=\lambda$
if we pass line through c then 
$\vec{r}=1\neq0$
SO $\lambda=3$ to satisfy eq 
Multiple choice direction cosines and direction ratios three dimensional geometry maths

If points (1,2), (3 , 5) and (0 , b ) are collinear the value of b is  

  1. $\dfrac{1}{2}$
  2. $\dfrac{7}{2}$

  3. 2

  4. -1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Area=\dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|$
As points are collinear , so area =0
$\therefore \dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|=0$
$\Rightarrow 5-b+3b-6=0$
$\Rightarrow=1=2b$
$\therefore b=\dfrac{1}{2}$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Three points whose position vectors are $x\bar{i}+y\bar{j}+z\bar{k}$, $\bar{i}+2\bar{j}$ and $-\bar{i}-\bar{j}$ are collinear, then relation between $x, y, z$ is?

  1. $x-2y=1, z=0$
  2. $z+y=1, z=0$
  3. $x-y=1, z=0$

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

For three points to be collinear, the vectors connecting them must be proportional. The vectors are P1(x, y, z), P2(1, 2, 0), and P3(-1, -1, 0). The vector P2P3 is (-2, -3, 0). The vector P1P2 is (1-x, 2-y, -z). For these to be parallel, the ratios of components must be equal, implying z=0 and a specific linear relationship between x and y that is not listed in A, B, or C.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(\alpha, - 1), (2, 1)$ and $(4, 5)$ are collinear, then find $\alpha $ by vector method.

  1. $4$
  2. $1$
  3. $8$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If there points are collinear then vectors from one to another will have scalar triple produced $0$.Point $\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$
$\left( 2-\alpha  \right) \hat { i } +2\hat { j } -\bar { A }$
$2\hat { i } +4\hat { j } -\bar { B }$
$\left( 4-\alpha  \right) \hat { i } +6\hat { j } -\bar { C }$
$ \bar { A } .\left( \bar { B } \times \bar { C }  \right) =0$
$\left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) \left( 2\hat { i } +4\hat { j }  \right) \times \left( \left( 4-\alpha  \right) \hat { i } +6\hat { j }  \right) \\ \left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k }  \right] =0$
$4\alpha =4$
 $\alpha =1$
Also the direction vector will be proportion
$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$
$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$
$\lambda=\dfrac{1}{2}$ as $2=4\lambda$
$2-\alpha=1$
$\therefore \alpha=1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$ are collinear, then  

  1. $k$ has only one real value
  2. $k$ has two real value
  3. $k$ has no real values
  4. $k$ has infinite number of real values
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As the $3$ point should be collinear area of triangle termed by then should be zero

considering $2$ direction to be $(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$
& $(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$

$(2\bar{b})\times (1-k)\bar{b}=0$

this will be for any value of $k$ as cross produced of $2$ linear vector $=0$


Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $A = (1,2,3) , B  = (2,10,1), Q$ are collinear points and $Q _{x}=-1$ then $Q _{z}$ is

  1. $-3$
  2. $7$
  3. $-14$
  4. $-7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since A(1, 2, 3), B(2, 10, 1), and Q(Qx, Qy, Qz) are collinear, the vector AB = (1, 8, -2) must be proportional to AQ = (Qx-1, Qy-2, Qz-3). Given Qx = -1, the x-component of AQ is -2. Since -2 is -2 times the x-component of AB, the other components must follow the same ratio: Qy-2 = -2(8) = -16 (Qy = -14) and Qz-3 = -2(-2) = 4 (Qz = 7).

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If points $\hat i + \hat j, \hat i - \hat j$ and $p \hat i + q \hat j + r \hat k$ are collinear, then

  1. $p = 1$
  2. $r = 0$
  3. $q \in R$
  4. $q \neq 1$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Points $A(\hat i + \hat j), B (\hat i - \hat j)$ and $C(p \hat i + q \hat j + r \hat k)$ are collinear
Now $\vec{AB} = - 2 \hat j$ and $\vec{BC} = (p -1) \hat i + (q - 1) \hat j + r k$
Vectors $\vec{AB}$ and $\vec{BC}$ must be collinear
$\Rightarrow p = 1,  r= 0 $ and $q \neq 1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $(12, 9, 2)$
  2. $(1, -1, -2)$
  3. $(5, -3, 4)$
  4. $(0, 0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know condition of collinearity,
$\dfrac{x-x _1}{x _2-x _1}=\dfrac{y-y _1}{y _2-y _1}=\dfrac{z-z _1}{z _2-z _1}$


We have points $(0,1,-2),(3,\lambda,-1) and (\mu,-3,-4)$
$\dfrac{x-0}{3-0}=\dfrac{y-1}{\lambda-1}=\dfrac{z+2}{-1+2}.........(1)$

Since $(\mu,-3,-4)$ is collinear , so it satisfy equation (1)
taking y and z coordinates,
$\dfrac{-3-1}{\lambda-1}=\dfrac{-4+2}{-1+2}$
$\lambda=3$

Now equation (1) becomes
$\dfrac{x-0}{3-0}=\dfrac{y-1}{3-1}=\dfrac{z+2}{-1+2}.........(2)$

Option (A) (12,9,2) satisfy equation (2).

Therefore option (A) is correct.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(-1, 3, 2), (-4, 2, -2)$ and $(5, 5, \lambda)$ are collinear, then $\lambda$ is equal to

  1. $-10$
  2. $5$
  3. $-5$
  4. $10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the points be $A(-1,3,2) B(-4,2,-2) C(5,5,\lambda)$
Direction ratio of $AB{=}(-3,-1,-4)$
If $A,B,C$ are collinear points then direction ratio of $AB$ and $BC$ must be proportional.
Direction ratio of $BC{=}(9,3,\lambda+2)$
$\therefore 9{=}\alpha (-3)$
$\therefore 3{=}\alpha(-1)$
$\therefore \lambda+2{=}\alpha(-4)$ and $\alpha{=} -3$
$\therefore \dfrac{\lambda+2}{-3}{=}-4$
$\therefore \lambda{=} 10$