Mathematics · Quantitative Aptitude

Coordinate Geometry

204 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 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)

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

$\Rightarrow 3bc-2ad=0$
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 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 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

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 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 $\alpha{=} -3$
$\therefore \dfrac{\lambda+2}{-3}{=}-4$
$\therefore \lambda{=} 10$

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

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$
$\therefore \dfrac{λ−1}{1}=\dfrac{−4}{4}=\dfrac{−3}{3}⇒λ−1=−1⇒λ=0$