Tag: constructions

Questions Related to constructions

The mid-point of a line segment is $(5,8)$. If one end point is $(3,5)$, find the second end point

  1. $(-7,11)$

  2. $(7,-11)$

  3. $(7,11)$

  4. $(-7,-11)$


Correct Option: A,C
Explanation:

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x

} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Let the other point be $ (x,y) $
So,
$=\left( \dfrac {3 + x }{ 2 } ,\dfrac { 5 + y }{ 2 }  \right) \quad =\quad (5,8)

$
$ => \frac {3 + x }{ 2 } = 5 $ and  $ \dfrac { 5 + y } {2} = 8 $
$ => x = 7 , y = 11 $
So, the second end point is $ (7,11) $

The vertices of a parallelogram are $(3, -2)$, $(4,0)$, $(6, -3)$ and $(5, -5)$. The diagonals intersect at the point M. The coordinates of the point M are:

  1. $\begin{pmatrix} \frac { 9 }{ 2 },-\frac { 5 }{ 2 } \end{pmatrix}$

  2. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 5 }{ 2 }\end{pmatrix}$

  3. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 3 }{ 2 }\end{pmatrix}$

  4. None of these


Correct Option: A
Explanation:

The diagonals of a parallelogram bisect each other. Hence M is the mid point of the vertices $ (3,-2) ; (6,-3) $ or of the vertices  $ (4,0) ; (5,-5) $

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is  calculated by the formula $ \left( \frac { { x } _{ 1 }+{x} _{ 2 } }{ 2 } ,\frac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Using this formula, mid point of $ (3, -2) , (6, -3) = \left( \frac { 3\quad +6 }{ 2 } ,\frac { -2-3 }{ 2 }  \right) =\left( \frac { 9 }{ 2 } ,\frac { -5 }{ 2 }  \right) $

Find the mid-point of AB where A and B are the points $(-5, 11)$ and $(7,3)$, respectively.

  1. $(1,7)$

  2. $(0,0)$

  3. $(1,0)$

  4. $(0,7)$


Correct Option: A
Explanation:

Midpoint of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is  calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x

} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Using this formula, mid point of AB $= \left( \dfrac { -5 + 7 }{ 2 } ,\dfrac { 11 + 3 }{ 2 } 

\right) = (1, 7) $

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.

  1. $(5,1)$

  2. $(1,1)$

  3. $(1,5)$

  4. $(1,1-)$


Correct Option: B
Explanation:

Let the vertices of the parallelogram be $A (-2,-1), B(1,0),  C(4,3),  D(1,2) $

The diagonals AC and BD would meet at the midpoint of AC and BD.

Midpoint of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{2 }) $ is  calculated by the formula $ \left( \cfrac { { x } _{ 1 }+{ x} _{ 2 } }{ 2 } ,\cfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Hence, 
mid point of AC $= \left( \cfrac { -2+4 }{ 2 } ,\cfrac { -1+3}{ 2 }  \right) = (1,1) $

The mid-point of a line is $(-4,-2)$ and one end of the line is $(-6,4)$. The co-ordinates of the other end are

  1. $(2,-8)$

  2. $(-2,8)$

  3. $(-2,-8)$

  4. $(2,8)$


Correct Option: C
Explanation:

The co-ordinate of the mid point of a line segment whose end points are $(x _1,y _) $ and $(x _2,y _2)$ are given by $(\cfrac{x _1+x _2}{2},\cfrac{y _1+y _2}{2})$

Substituting $(x _1,y _1)=(-6,4)$ and mid-point $(-4,-2)$.
$\Rightarrow (-4,-2)=(\cfrac{-6+x _2}{2},\cfrac{4+y _2}{2})$
Equating the $x-$coordinates
$\Rightarrow -4=\cfrac{-6+x _2}{2}$
$\Rightarrow -4\times 2=-6+x _2$
$\Rightarrow x _2=6-8$
$\Rightarrow x _2=-2$
Equating the $y-$coordinates

$\Rightarrow -2=\cfrac{4+y _2}{2}$
$\Rightarrow -2\times 2=4+y _2$
$\Rightarrow -y _2=4+4$
$\Rightarrow y _2=-8$

The coordinates of the other end are $(-2,-8)$.

The end points of a diagonal of a parallelogram are $(1, 3)$ and $(5, 7)$, then the mid-point of the other diagonal is ..........

  1. $(1, 7)$

  2. $(3, 5)$

  3. $(5, 3)$

  4. $(7, 1)$


Correct Option: B
Explanation:

Given end points of diagonal of parallelogram $(1,3)$ and $(5,7)$

We know that the mid point of two point $(x _1,y _1) \ and \  (x _2,y _2)$ are
$\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}$
$\Rightarrow \dfrac{1+5}{2},\dfrac{3+7}{2}$
$\Rightarrow( 3,5)$

Calculate mid point of $A(5,\,3)$ and $B(9,\,8)$

  1. $\dfrac{11}{2},\,7$

  2. $7,\,\dfrac{11}{2}$

  3. $7,\,11$

  4. $14,\,11$


Correct Option: B
Explanation:

The mid point of line segment joining $(x _1,\,y _1)$ and $(x _2,\,y _2)$ is $\begin{pmatrix}\dfrac{x _1+x _2}{2},\,\dfrac{y _1+y _2}{2}\end{pmatrix}$
Mid point of AB $\Rightarrow\begin{pmatrix}\dfrac{5+9}{2},\,\dfrac{3+8}{2}\end{pmatrix}$
$\Rightarrow\begin{pmatrix}7,\,\dfrac{11}{2}\end{pmatrix}$

Three vertices of rhombus taken in order are $(2, -1), (3, 4)$ and $(-2, 3)$. Find the fourth vertex.

  1. $(1, 2)$

  2. $(-3, -2)$

  3. $(3, 2)$

  4. None of these


Correct Option: B
Explanation:

Let M be the mid point of one diagonal formed by (2, -1) and (-2, 3) then using mid-point method;
Coordinate of $M \left (\dfrac {2 - 2}{2}, \dfrac {-1 + 3}{2}\right ) = (0, 1)$
$\Rightarrow \left (\dfrac {x + 3}{2}, \dfrac {y + 4}{2}\right ) = (0, 1)$
$\Rightarrow x = -3$ and $y = -2$
$\therefore (-3, -2)$ is coordinate of $D$

What is the midpoints between the coordinates $(-1, 2)$ and $(-1, -6)$?

  1. $\left(1, 2\right)$

  2. $\left(-1, 2\right)$

  3. $\left(-1, -2\right)$

  4. $\left(1, -2\right)$


Correct Option: C
Explanation:

We know the midpoint formula between two points.
$\left(\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}\right)$
Substitute the values, we get
$=$ $\left(\dfrac{-1-1}{2},\dfrac{2-6}{2}\right)$
$=$ $\left(\dfrac{-2}{2},\dfrac{-4}{2}\right)$
Therefore, midpoint is $\left(-1, -2\right)$

What is the midpoints between the coordinates $(0, -6)$ and $(4, -4)$?

  1. $\left(-2, -5\right)$

  2. $\left(2, 5\right)$

  3. $\left(2, -4\right)$

  4. $\left(2, -5\right)$


Correct Option: D
Explanation:

We know the midpoint formula between two points.
$\left(\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}\right)$
Substitute the values, we get
$=$ $\left(\dfrac{0+4}{2},\dfrac{-6-4}{2}\right)$
$=$ $\left(\dfrac{4}{2},\dfrac{-10}{2}\right)$
Therefore, midpoint is $\left(2, -5\right)$.