Mathematics

Coordinate Geometry and Graphs

77 Questions

Coordinate geometry involves plotting linear equations, analyzing graphical data, and understanding planar graphs using vertices, edges, and faces. These concepts are essential for calculating intersection points and interpreting graphical information. Such questions frequently appear in engineering, state, and civil services examinations.

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Coordinate Geometry and Graphs Questions

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

The point of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

  1. lie on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  2. lie on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  3. lie on a circle centered at $\displaystyle \left( 8 ,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 } } $

  4. are not concyclic
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If ${S} _{1}=0$ and ${S} _{2}=0$ are the equations, then, $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${S} _{1}=0$ and ${S} _{2}=0.$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$   ...(1)
For it to be a circle, choose $\lambda$ such that the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal: $\Rightarrow \lambda+4=2\lambda+2$
$\therefore \lambda=2$
This gives the equation of the circle as $6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$  {(using (1))}
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x-6y+\frac { 27 }{ 2 } =0$ 
Its centre is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

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

The line $x+y=1$ meets the lines represented by the equation $y^{3}-xy^{2}-14x^{2}y+24x^{3}=0$ at the points $A, B, C$. If $O$ is the origin, then $OA^{2}+OB^{2}+OC^{2}$ is equal to

  1. $\dfrac{22}9$
  2. $\dfrac{85}{72}$
  3. $\dfrac{181}{72}$
  4. $\dfrac{221}{72}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

X-coordinate of the points are given by the roots of the equation


$24{ x }^{ 3 }+14{ x }^{ 2 }\left( x-1 \right) -x{ \left( x-1 \right)  }^{ 2 }-{ \left( x-1 \right)  }^{ 3 }=0\ \Rightarrow 36{ x }^{ 3 }-9{ x }^{ 2 }-4x+1=0\ \Rightarrow \left( 3x-1 \right) \left( 3x+1 \right) \left( 4x-1 \right) =0\ \Rightarrow x=\cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 4 } $

$\Rightarrow A\left( \cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 }  \right) ,B\left( -\cfrac { 1 }{ 3 } ,\cfrac { 4 }{ 3 }  \right) $ and $C\left( \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 4 }  \right) $

Hence,

${ OA }^{ 2 }+{ OB }^{ 2 }+{ OC }^{ 2 }=\cfrac { 1 }{ 9 } +\cfrac { 4 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 16 }{ 9 } +\cfrac { 1 }{ 16 } +\cfrac { 9 }{ 16 } =\cfrac { 221 }{ 72 } $

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

The points of intersection of the two ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$.

  1. lie on a circle centred at $\left(\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  2. lie on a circle centred at $\left(-\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  3. lie on a circle centred at $(8, 9)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  4. are not cyclic.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of any curve passing through the intersection of the given ellipse is
   $4x^{2}+2y^{2}-20x-12y+35+\lambda \left ( x^{2}+2y^{2}-6x-12y+23 \right )=0$
which represents a circle is
   $4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
   $6x^{2}+6y^{2}-32x-36y+81=0$
$\Rightarrow $   $\displaystyle x^{2}+y^{2}-\left ( \frac{16}{3} \right )x-6y+\frac{81}{6}=0$
centre of the circle is $\left(\dfrac83, 3\right)$
and the radius is $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

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

The set of values of $c$ so that the equations $\displaystyle y=\left | x \right |+c: : and: : x^{2}+y^{2}-8\left | x \right |-9=0 $ have no solution is

  1. $\displaystyle \left ( -\infty ,-3 \right )\cup \left ( 3,\infty \right )$
  2. $(-3, 3)$
  3. $\displaystyle \left ( -\infty ,-5\sqrt{2} \right )\cup \left ( 5\sqrt{2},\infty \right )$
  4. $\displaystyle \left ( -\infty ,-4-5\sqrt{2} \right )\cup \left ( 5\sqrt{2}-4,\infty \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation 

$y=\left |x\right |+c$---(1)
$x^2+y^2-8\left| x\right |-9=0$
From equation (1) and (2)
$x^2+(\left | x\right |+c)^2-8\left |x\right |-9=0$
when $x>0$
$x^2+(x+c)^2-8x-9=0$
$x^2+x^2+c^2+2cx-8x-9=0$
$2x^2+x(2c-8)+c^2-9=0$
For no solution 
$D<0$
$(2c-8)^2-4\times 2 (c^2-9)<0$
$4c^2+64-32c-8c^2+72<0$
$-4c^2-32c+136<0$
$c^2+8c-34>0$
$c=\dfrac{-8\pm\sqrt{64+136}}{2}$

$c=\dfrac{-8\pm\sqrt{200}}{2}$

$c=-4\pm 5\sqrt{2}$
C has root $c=-4\pm5\sqrt{2}$
Hence for no solution c has all value excluding it's roots  
$c\epsilon(-\infty,-4-5\sqrt{2})\cup(5\sqrt{2}-4,\infty)$

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

The points of intersection of the two ellipses ${ x }^{ 2 }+2{ y }^{ 2 }-6x-12y+23=0$ and $4{ x }^{ 2 }+2{ y }^{ 2 }-20x-12y+35=0$

  1. lies on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  2. lies on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  3. lies on a circle centered at $\displaystyle \left( 8,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 } } $

  4. are not cyclic

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$ are the equations, then $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$
For it to be a circle, choose $\lambda$ such that the coefficients of ${ x }^{ 2 }$ and ${ y }^{ 2 }$ are equal:
$\Rightarrow \lambda +4=2\lambda +2\Rightarrow \lambda =2$
This gives the equation of the circle as
$\displaystyle 6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$    (Using (1))
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x+6y+\frac { 27 }{ 2 } =0$
Its center is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

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

How many points of intersection are between the graphs of the equations $x^2+ y^2 = 7$ and $x^2- y^2 = 1$?

  1. $0$
  2. $1$
  3. $2$$
  4. $3$
  5. $4$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Given ${x}^{2}+{y}^{2}=7$ and ${x}^{2}-{y}^{2}=1$
Add two equations, we get $2{x}^{2}=8$ , which implies ${x}^{2}=4$
Therefore $x = \pm2$ , we get $y=\pm \sqrt3$
So, number of solutions is $4$.

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.

  1. $(7,6)$
  2. $(-7,-6)$
  3. $(7,-6)$
  4. $(-7,6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

For $\theta=180^{0}$ , the transformation matrix will be $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$
So the image of point $(-7,6)$ is $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} -7 \ 6 \end{bmatrix}=\begin{bmatrix} 7 \ -6 \end{bmatrix}$
Therefore the correct option is $C$

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The equation of image of pair of lines $y=|x-1|$ with respect to y-axis is 

  1. ${x^2} - {y^2} - 2x + 1 = 0$
  2. ${x^2} - {y^2} - 4x + 4 = 0$
  3. $4{x^2} - 4x - {y^2} + 1 = 0$
  4. ${x^2} - {y^2} + 2x + 1 = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have $y=|x-1|$


$\Rightarrow y^2=(x-1)^2$

Change $x$ by $-x$, then the required image is 

$y^2=(-x-1)^2$

$\Rightarrow y^2=x^2+2x+1$

$\Rightarrow x^2-y^2+2x+1=0$

Multiple choice business economics and quantitative methods government budget and economy consumer's budget public finance indifference curve

Slope of budget line is Indicated by:

  1. $\displaystyle \frac{P _X}{P _Y}$
  2. $\displaystyle \frac{P _Y}{P _X}$
  3. $P _X = P _Y$

  4. all of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The slope of the budget line is determined by the ratio of the prices of the two goods. It represents the rate at which the market allows a consumer to trade one good for another.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The graph of the equation $4y^2 + x^2= 25$ is

  1. a circle

  2. an ellipse

  3. a hyperbola

  4. a parabola

  5. a straight line
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $4{y}^{2}+{x}^{2}=25$

$\Rightarrow \dfrac { { y }^{ 2 } }{ 25/4 } +\dfrac { { x }^{ 2 } }{ 25 } =1$
It is in the form of ellipse $\left (\dfrac { { y }^{ 2 } }{ {a}^{2} } +\dfrac { { x }^{ 2 } }{ {b}^{2} } =1\right)$
So, the correct answer is option $B$.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A parabola $y = ax^2 + bx + c$ crosses the x-axis at $(\alpha, 0)$ $(\beta, 0)$ both to the right of the origin. A circle also passes through these two points. The length of the tangent from the origin to the circle is

  1. $\displaystyle \sqrt{\frac{bc}{a}}$
  2. $ac^2$
  3. $\displaystyle \frac{b}{a}$
  4. $\displaystyle \sqrt{\frac{c}{a}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$OT$ is a tangent and $OAB$ is a secant 


we know that

$OT^2 =OA.OB$

         $=\alpha\beta$

         $=\dfrac{c}{a}$ (Since $\alpha,\beta $ are the roots of $y=ax^2+bx+c$)

$\Rightarrow OT=\sqrt{\dfrac{c}{a}}$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

For what positive value(s) of K will the graph of the equation $2x + y = K$ be tangent to the graph of the equation $x^2+ y^2= 45$?

  1. 5

  2. 10

  3. 15

  4. 20

  5. 25
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  • The radius of circle is $\sqrt{45} = 3\sqrt5$ , center of circle is $(0,0)$
  • For the equation to be tangent to circle , the distance from center of circle to given line must be equal to radius of circle
  • So we get $k/\sqrt5 = 3\sqrt5$ , which gives $k=15$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If a pair of perpendicular straight lines drawn through the origin forms an isosceles triangle with the line $2x+3y=6$, then area of the triangle so formed is?

  1. $36/13$
  2. $12/17$
  3. $13/5$
  4. $17/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the lines are perpendicular and form an isosceles triangle the other two angles must be $45^\circ$

Let the slope of the line be m
$tan 45^\circ = \Bigg|\cfrac{-\cfrac{2}{3}-m}{1-\cfrac{2m}{3}}\Bigg| = 1$
$m = -5$ and the other line slope =$\cfrac{1}{5}$
Lines 
$y +5x= 0$ and $5y =x$
Intersection points $(0,0)$ , $(-\cfrac{6}{13} , \cfrac{30}{13})$ and $(\cfrac{30}{13} , \cfrac{6}{13})$
Perpendicular distance from origin to line $2x+3y=6$ is $\cfrac{|0+0-6|}{\sqrt{2^2+3^2}} = \cfrac{6}{\sqrt{13}}$
Distance between the points are $\sqrt{\Bigg(\cfrac{36}{13}\Bigg)^2+\Bigg(\cfrac{24}{13}\Bigg)^2} = \cfrac{\sqrt{1872}}{13} = \cfrac{12\sqrt{13}}{13}$
Area = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$ 

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $(0.-1), (2, 1) and (0, 3)$

  1. $4$
  2. $8$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x-1(y _2-y _3)+x _2(y _3+y _1)+x _3(y _1-y _2)]\\=2[0+2(3-1)+0]=8sq\>unit$