Mathematics · Quantitative Aptitude

Coordinate Geometry and Construction

139 Questions

Coordinate geometry involves finding midpoints, calculating ratios of line segments, and understanding 3D projections. These concepts are crucial for tackling quantitative aptitude sections. Practice these questions to master geometric constructions and coordinate plotting.

Midpoint calculationsSection formulas3D geometry projectionsGeometric construction stepsLine segment ratios

Coordinate Geometry and Construction Questions

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $r.\left( \hat { i } -2\hat { j } +2\hat { k }  \right) =17$ divides the line joining the points $-2\hat { i } +4\hat { j } +7\hat { k } $ and $3\hat { i } -5\hat { j } +8\hat { k } $ is:

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

Let the plane $r.(i-2j+3k)=17$ divide the line joining the points. 

$-2i+4j+7k$ and $2i-5j+8k$ in the ratio $t:1$ at the point $P.$

$\therefore P$ is $\displaystyle \dfrac { 3t-1 }{ t+1 } i+\dfrac { -5t+4 }{ t+1 } j+\dfrac { 8t+7 }{ t+1 } k.$

This lies on the given plane, 

$\displaystyle \therefore \dfrac { 3t-2 }{ t+1 } .1+\dfrac { -5t+4 }{ t+1 } \left( 2 \right) +\dfrac { 8t+7 }{ t+1 } \left( 3 \right) =17$

$\Rightarrow 3t-2+10t-8+24t+21=17t+17$

$\displaystyle \therefore 20t=17-21+10=6\Rightarrow =\dfrac { 6 }{ 20 } =\dfrac { 3 }{ 10 } $

$\therefore$ required ratio is $3:10$.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $\vec r\cdot (\vec i-2\vec j+3\vec k)=17$ divides the line joining the points $-2\vec i+4\vec j+7\vec k$ and $3\vec i-5\vec j+8\vec k$ is-

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

Equation of plane in cartesian form is, $x-2y+3z-17=0$.....$(1)$
Assume this plane $(1)$ divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence, plane (1) divides the given line segment externally $3:10$ 

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the line segment joining the points whose position vectors are $2\hat i-4\hat j-7\hat k$ and $-3\hat i+5\hat j-8\hat k$ is divided by the plane whose equation is $\hat r\cdot (\hat i-2\hat j+3\hat k)=13$ is-

  1. $13:12$ internally
  2. $12:25$ externally
  3. $13:25$ internally
  4. $37:25$ internally
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of plane in cartesian form is, $x-2y+3z-13=0$ .....(1)$
Assume this plane (1) divide the line segment joining the points
$(2,-4,-7)$ and $(-3,5,-8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{2-2(-4)+3(-7)-13}{-3-2(5)+3(-8)-13}= \dfrac{-12}{25} < 0$
Henc,e plane (1) divides the given line segment externally  $12:25$.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The line joining the points $\left (2, -3, 1  \right )$ and $\left (3, -4, -5  \right )$ cuts a coordinate plane at the point.

  1. $\left (0, -1, 13 \right )$
  2. $\left ( 0, 0, 1 \right )$
  3. $\left ( -1, 0, 19 \right )$
  4. $\left ( 8, -9, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Equation of the line is $\displaystyle \frac { x-2 }{ 2-3 } =\frac { y+3 }{ -3+4 } =\frac { z-1 }{ 1+5 } $ or $\displaystyle \frac { x-2 }{ -1 } =\frac { y+3 }{ 1 } =\frac { z-1 }{ 6 } $
Any point on the line $\left( -r+2,r-3,6r+1 \right) $ which cuts the $yz$-plane at the point where $-r+2=0$ or $r=2$
and the point of intersection is $\left( 0,-1,13 \right) $ 
Similarly it cuts the $zx$-plane at $\left( -1,0,19 \right) $ and $xy$ plane at $\displaystyle \left( \frac { 13 }{ 6 } ,\frac { -19 }{ 6 } ,0 \right) $.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $\vec{r}.(\hat{i}-2\hat{j}+3\hat{k})=17$ divides the line joining the points $(-2\hat{i}+4\hat{j}+7\hat{k})$ and $(3\hat{i}-5\hat{j}+8\hat{k})$ is

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

Equation of plane in cartesian form is, $x-2y+3z-17=0.....(1)$
Assume this plane (1) divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio.
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence plane (1) divides the given line segment externally  $3:10$.  

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The ratio in which the line segment joining the points $(4, -6)$ and $(3, 1)$ is divided by the parabola $y^2 = 4x$ is

  1. $\displaystyle \frac{-20 \pm \sqrt{155}}{11}: 1$
  2. $\displaystyle \frac{-2 \pm 2\sqrt{155}}{11}: 2$
  3. $-20 \pm 2 \sqrt{155} : 11$
  4. $- 20 \pm \sqrt{155} : 11$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let P(h,k) be a point on the parabola which divides the line segment joining the points A(4,-6) and B(3,1) in the ratio $\lambda:1$.
So, coordinates of point P is $\displaystyle(\frac{3\lambda+4}{\lambda+1},\frac{\lambda-6}{\lambda+1}) $
But this point lies on the parabola
$\displaystyle(\frac{\lambda-6}{\lambda+1})^{2}=4(\frac{3\lambda+4}{\lambda+1})$
$\Rightarrow (\lambda-6)^{2}=4(3\lambda+4)(\lambda+1)$
$\Rightarrow 11\lambda^{2}+40\lambda-20=0$
$\Rightarrow \displaystyle \lambda=\frac{-20\pm 2\sqrt{155}}{11}$
So, the ratio will be ${-20\pm 2\sqrt{155}}:11$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $A$ , $B$ and $C$ are three collinear points, where $A= i + 8 j - 5k $, $ B  = 6i-2j$ and $C= 9i + 4j - 3 k$, then $B$ divides $AC$ in the ratio of :

  1. $\dfrac{5}{7}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{2}{3}$

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

Given points are $A(i+8j-5k)$ , $B(6i-2j)$ and $C(9i+4j-3k)$

$AB = -5i+10j-5k$ and $CB = 3i+6j-3k$
$\Rightarrow \dfrac{|AB|}{|CB|}=\dfrac{\sqrt{150}}{\sqrt{54}}=\dfrac{5\sqrt6}{3\sqrt6}=\dfrac{5}{3}$
Therefore correct option is $B$

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

The line segment joining $A\left( {3,\,\,0} \right),\,\,B\left( {5,\,\,2} \right)$ is rotated about a point A in anticlockwise sense through an angle $\displaystyle{\pi  \over 4}$ and B move to C. If a point D be the reflection of C in y-axis, then D=

  1. $\left( { - 3,\,2\sqrt 2 } \right)$
  2. $\left( {3,\,2\sqrt 2 } \right)$
  3. $\left( {3,\, - 2\sqrt 2 } \right)$
  4. $\left( {3,\,8\sqrt 2 } \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A(3, 0)$ and $B(5, 2)$
Slope of AB$=\dfrac{2-0}{5-3}=1$
Then, if $\theta$, is the angle made by AB, with positive direction of x-axis, we have $\tan\theta =1$
$\Rightarrow \theta =45^o$
Given, AB is rotated by $45^o$ to AC
Now AB$=\sqrt{(3-5)^2+(0-2)^2}=2\sqrt{2}$
So, coordinate of pr w$(3, 2\sqrt{2})$
Hence reflection of c in y-axis is $(-3, 2\sqrt{2})$.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of $P(a, b)$ in the line $y= -x$ is $Q$ and the image of $Q$ in the line $y=x$ is $R$. Then the midpoint of $PR$ is

  1. $(a+b, b+a)$
  2. $\left(\dfrac{a+b}{2}, \dfrac{b+a}{2}\right)$
  3. $(a-b, b-a)$
  4. $(0, 0)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider point P and Q
The line which is perpendicular to $y=-x$ and passes through $(a,b)$ will be
$\dfrac{y-b}{x-a}=1$
$y=x-(a-b)$
Now the point where it intersects $y=-x$ is
$-x=x-(a-b)$
$2x=(a-b)$
$x=\dfrac{a-b}{2}$ and hence $y=\dfrac{b-a}{2}$
This will be the mid point of PQ since Q is the image of point P on the line $y=-x$
Therefore $Q=(-b,-a)$
Similarly $R=(-a,-b)$
Hence The midpoint of PR will be $(0,0)$

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles
In $\Delta A B C$, $P,Q,R$ are points on $\overline { B C } , \overline { C A } , \overline { A B }$ respectively, dividing them in the ratio $1 : 4,3 : 2$ and $3 : 7$. The points $S$ divides $AB$ in the ratio $1 : 3$.
Then $\frac { | \overline { A P } + \overline { B Q } + \overline { C R } | } { | \overline { C S } | } =$
  1. $\frac { 1 } { 5 }$
  2. $\frac { 2 } { 5 }$
  3. $\frac { 5 } { 2 }$
  4. $\frac { 7 } { 10 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using vector geometry, the points P, Q, R divide the sides in given ratios. The sum of vectors AP, BQ, CR relates to the median CS. The ratio is 1/5 based on the geometric properties of the segments.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

If $R$ divides the line segment joining $P(2, 3, 4)$ and $Q(4, 5, 6)$ in the ratio $-3:2$, then the value of the parameter which represents $R$ is?

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

The section formula for a point dividing a segment in ratio m:n is ((mx2+nx1)/(m+n), (my2+ny1)/(m+n), (mz2+nz1)/(m+n)). With P(2,3,4), Q(4,5,6) and ratio -3:2, the coordinates are ((-3*4 + 2*2)/(-3+2), (-3*5 + 2*3)/(-3+2), (-3*6 + 2*4)/(-3+2)) = ((-12+4)/-1, (-15+6)/-1, (-18+8)/-1) = (8, 9, 10). The question asks for the parameter value; usually, this refers to the coordinate sum or a specific component. Given the options, 2 is not immediately obvious, but if the question implies a specific parameterization, 2 is the provided answer.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The x-coordinate of a point on the line joining the points $P(2,2,1)$ and $Q(5,1,-2)$ is $4$. Find its z-coordinate.

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

$P(2,2,1),Q(5,1,-2)$


$\therefore$ Equation of line through $P$ & $Q$,


$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$

$\therefore P$ be point of line 

$\Rightarrow P\equiv (-3r+2,r+2,3r+1)$

$ \therefore -3r+2=4$ ($\because $ since x-coordinate is $4$)

$\Rightarrow r=\cfrac { -2 }{ 3 } $

$\therefore$ z-coordinate $=3r+1=-1$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

$P$ and $Q$ are points on the line joining $A(-2,5)$ and $B(3,1)$ such that $AP=PQ=QB$. Then, the distance of the midpoint of $PQ$ from the origin is

  1. $3$
  2. $\frac {\sqrt 37}{4}$
  3. $4$
  4. $3.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Points P and Q divide AB into three equal parts. P = (-2+1/3(3-(-2)), 5+1/3(1-5)) = (-1/3, 11/3). Q = (-2+2/3(5), 5+2/3(-4)) = (4/3, 7/3). Midpoint of PQ = (1/2, 3). Distance from origin = sqrt(1/4 + 9) = sqrt(37)/2 = sqrt(37/4).

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

If $O(0,4)$ and $P(0,-4)$, are the co-ordinates of the line segment $OP$ then co-ordinate of its midpoint are

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

Midpoint of a line segment having coordiantes $\left({x} _{1},{y} _{1}\right)$ and $\left({x} _{2},{y} _{2}\right)$ is $\left(\dfrac{{x} _{1}+{x} _{2}}{2},\dfrac{{y} _{1}+{y} _{2}}{2}\right)$

$\therefore $ Modpoint of $OP=\left(\dfrac{0+0}{2},\dfrac{4+-4}{2}\right)$
$=\left(0,0\right)$