Tag: constructions

Questions Related to constructions

If $\left (\dfrac {a}{3}, 4\right )$ is the midpoint of the line segment joining $A (-6, 5)$ and $B(-2, 3)$, find $a$.

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

  1. $-4$

  2. $-12$

  3. $12$

  4. $-6$

Show answer
Correct Option: B
Explanation:

Given the point A(-6,5) and B(-2,3) $\left( \dfrac { a }{ 3 } ,4 \right) $ is the middle of AB

$\Rightarrow \left( \dfrac { -2-6 }{ 2 } ,\dfrac { 3+5 }{ 2 }  \right) =\left( \dfrac { a }{ 3 } ,4 \right) \ \Rightarrow \dfrac { -8 }{ 2 } =\dfrac { a }{ 3 } \ \therefore a=-\dfrac { 24 }{ 2 } =-12\ $

$A(-3,2)$ and $B(5,4)$ are the end points of a line segment, find the coordinates of the midpoints of the line segment.

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

  1. $(1,3)$

  2. $(3,3)$

  3. $(1,1)$

  4. $(3,1)$

Show answer
Correct Option: A
Explanation:

Since $A(-3,2)\equiv(x _1,y _1)$ and $B(5,4)\equiv(x _2,y _2)$ are the end points of a line segment.


Therefore, the coordinates of the midpoints of the line segment is given by:


$(x,y)=\left( \dfrac { { x } _{ 1 }+{ x } _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+{ y } _{ 2 } }{ 2 }  \right)$

$ \\ \Rightarrow (x,y)=\left( \dfrac { -3+5 }{ 2 } ,\dfrac { 2+4 }{ 2 }  \right) \quad $

$\\ \Rightarrow (x,y)=\left( \dfrac { 2 }{ 2 } ,\dfrac { 6 }{ 2 }  \right) =\left( 1,3 \right)$ 

If $(-2,3), (4,-3), (4,5)$ are mid-points of the sides of a triangle, find the coordinates of the centroid of the triangle formed by these mid-points.

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

  1. $\left (3,\dfrac43 \right )$

  2. $\left (2,\dfrac43 \right )$

  3. $\left (2,\dfrac53 \right )$

  4. $\left (3,\dfrac53\right )$

Show answer
Correct Option: C
Explanation:

Let the given vertices of a triangle be A$(-2,3)$ and B $(4,-3)$ let the third vertex be C $(4,5)$.


Let Centroid be $G= \left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$


$\Rightarrow G$ = $\left (\dfrac{-2 +4+4 }{3} ,\dfrac{3-3+5}{3}\right )$

$\Rightarrow G$ = $\left (\dfrac{6 }{3} , \dfrac{5}{3}\right )$

$\Rightarrow G$ = $\left (2 , \dfrac {5}{3}\right )$.

Find the third vertex of a triangle, if two of its vertices are $(-3,1), (0,-2)$ and centroid is at the origin.

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

  1. $(3,4)$

  2. $(2,1)$

  3. $(3,2)$

  4. $(3,1)$

Show answer
Correct Option: D
Explanation:

Let the third vertex be $(a,b)\equiv(x _1,y _1),(-3,1)\equiv(x _2,y _2), (0,-2)\equiv(x _3,y _3)$


Centroid of triangle is $(0,0)$

$G =\left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$

$\Rightarrow \left( \dfrac { a-3+0 }{ 3 } ,\dfrac { b+1-2 }{ 3 }  \right) =(0,0)$

$\Rightarrow\dfrac{a-3}3=0$ and $\dfrac{b-1}3=0$

$\Rightarrow a=3,b=1$

So the third vertex is $(3,1)$

Find the midpoint of the line segment joining the points  $(1,-1)$ and $(-5,-3)$

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

  1. $(-2,1)$

  2. $(2,1)$

  3. $(-2,-1)$

  4. None of these

Show answer
Correct Option: A
Explanation:
Take $(x _1,y _1)=(1,-1)$ and $(x _2,y _2)=(-5,-3)$.

By midpoint theorem:

$x=\dfrac{x _1+x _2}{2}$ and $y=\dfrac{y _1+y _2}{2}$

Hence, $x=\dfrac{1+(-5)}{2}=-2$ and $y=\dfrac{-1-(-3)}{2}=1$.

So, $(x,y)=(-2,1)$.

The centre of a circle is at $(-6,4)$. If one end of a diameter of the circle is at the origin, then find the other end.

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

  1. $(-12,8)$

  2. $(-12,-8)$

  3. $(12,8)$

  4. None of these

Show answer
Correct Option: A
Explanation:

Take coordinate of other end of diameter as $P(x,y)$.


since, origin divides the diameter of a circle in $2$ equal parts, hence origin will be mid-point of $(0,0)$ and $(x,y)$.


By midpoint theorem:

$x=\dfrac{x _1+x _2}{2}$ and $y=\dfrac{y _1+y _2}{2}$

$=>-6=\dfrac{0+x}{2}=\dfrac{x}{2}$

$=>x=-12$

And,

$=>4=\dfrac{y+0}{2}=\dfrac{y}{2}$

$=>y=8$.

So, $P(x,y)=(-12,8)$

Find the mid point of (3,8) and (9,4).

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

  1. $(5,6)$

  2. $(6,6)$

  3. $(4,4)$

  4. None of the above

Show answer
Correct Option: B
Explanation:

Midpoint formula is given by $\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2} \right)$

So midpoint of $(3,8)$ and $(9,4)$ is $=\left(\dfrac{3+9}{2},\dfrac{8+4}{2} \right)=(6,6)$

Find the mid point of $(4,6)$ and $(2,-6)$.

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

  1. $(3,4)$

  2. $(2,-2)$

  3. $(3,0)$

  4. None of the above

Show answer
Correct Option: C
Explanation:

We know that end points of a line segment is $(a,b)$ and $(c,d)$, then the midpoint of the line segment has the coordinates:

$\dfrac{a+c}{2},\dfrac{b+d}{2}$
Then mid point of line segment $(4,6)$ and $(2,-6)$ is 

$\dfrac{4+2}{2},\dfrac{6-6}{2}$
$\Rightarrow \dfrac{6}{2},\dfrac{0}{2}$
$\Rightarrow (3,0)$

If mid point of the line segment joining (2a, 4) and (-2, 3b) is  (1, 2a + 1), then the values of  a and b are given by

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

  1. $a = 2, b = - 2$

  2. $a = b = 2$

  3. $a= 1 = b$

  4. $a= -2, b = 2$

Show answer
Correct Option: B
Explanation:
Midpoint of any two points $(a.b)$ and $(c,d)$ is given by
$M=\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right)$
Given points are $(2a, 4)$ and $(-2,3b)$ 
$\therefore M=\left(\dfrac{2a-2}{2},\dfrac{4+3b}{2}\right)$
$\implies (1,2a+1)=\left(a-1,\dfrac{4+3b}{2}\right)$
$\implies a-1=1$ and $\dfrac{4+3b}{2}=2a+1$
$\implies a=2$ and $4a-3b=2$
$\implies a=2, b=2$
Hence, $a=b=2$.

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points $(-2,-1),(1,0),(4,3)$ and $(1,2)$ meet.

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

  1. $(1,1)$

  2. $(1,3)$

  3. $(5,1)$

  4. None of these

Show answer
Correct Option: A
Explanation:

The vertices of parallelogram in order are $A(-2,-1), B(1,0), C(4,3), D(1,2)$.


So the diagonals will be $AC$ and $BD$.

Since the diagonals of a parallelogram bisects each other, so mid-point of 

$AC$ or $BD$ will be intersection point of diagonals.

Hence by mid-point theorem, mid-point of $AC$ is

$A(-2,-1) \  and \ C(4,3)$

$x=\dfrac{-2+4}{2}=1$ and $y=\dfrac{-1+3}{2}=1$.

so $(1,1)$ is required point.