Tag: maths

Questions Related to maths

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

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

  1. $(2,-8)$

  2. $(-2,8)$

  3. $(-2,-8)$

  4. $(2,8)$

Show answer
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 ..........

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

  1. $(1, 7)$

  2. $(3, 5)$

  3. $(5, 3)$

  4. $(7, 1)$

Show answer
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 $B(9,\,8)$

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

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

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

  3. $7,\,11$

  4. $14,\,11$

Show answer
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.

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

  1. $(1, 2)$

  2. $(-3, -2)$

  3. $(3, 2)$

  4. None of these

Show answer
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 $(-1, -6)$?

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

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

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

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

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

Show answer
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 $(4, -4)$?

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

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

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

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

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

Show answer
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 $\left(2, -5\right)$.

Find the centre of circle, if the coordinates of two ends of diameter are $(-1, 7)$ and $(11, 5)$

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

  1. $(5, 6)$

  2. $(-3, 2)$

  3. $(10, 12)$

  4. $(12, 2)$

Show answer
Correct Option: A
Explanation:

$x = \dfrac {11 - 1}{2} = 5$ and $y = \left (\dfrac {5 + 7}{2}\right )$ $= 6$ (hint: using mid-point formula for centre).
$\therefore (5, 6)$ is coordinate of centre

Find the midpoint between the coordinates $(9, 3)$ and $(1, 1)$.

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

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

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

  3. $\left(5, 1\right)$

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

Show answer
Correct Option: A
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
$=$ $\left(\dfrac{9+1}{2},\dfrac{3+1}{2}\right)$
$=$ $\left(\dfrac{10}{2},\dfrac{4}{2}\right)$
Therefore, midpoint is $\left(5, 2\right)$

Find the value of $k$, so that $(2, 1)$ is the midpoint between $(1, k)$ and $(3, 1)$.

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
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
$\left(\dfrac{1+3}{2},\dfrac{k+1}{2}\right)$ $=$ $\left(2, 1\right)$
On equating, we get
$\left(\dfrac{k+1}{2}\right)$ $= 1$
$\Rightarrow k + 1 = 2$
$\Rightarrow k = 2 - 1$
$\Rightarrow k = 1$

Find the midpoints between the coordinates $(2, 3)$ and $(1, 0)$

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

  1. $\left(\dfrac{1}{2},\dfrac{3}{2}\right)$

  2. $\left(\dfrac{3}{2},\dfrac{1}{2}\right)$

  3. $\left(\dfrac{3}{2},\dfrac{4}{2}\right)$

  4. $\left(\dfrac{3}{2},\dfrac{3}{2}\right)$

Show answer
Correct Option: D
Explanation:

Midpoint between the coordinates $(x _1,y _1)$ and $(x _2,y _2)$ is:

$\left( \dfrac { x _{ 1 }+x _{ 2 } }{ 2 } ,\dfrac { y _{ 1 }+y _{ 2 } }{ 2 }  \right)$
Therefore, the midpoint between the coordinates $(2,3)$ and $(1,0)$ is: 
$\left( \dfrac { 2+1 }{ 2 } ,\dfrac { 3+0 }{ 2 }  \right) =\left( \dfrac { 3 }{ 2 } ,\dfrac { 3 }{ 2 }  \right)$
Therefore, midpoint is $\left (\dfrac {3}{2}, \dfrac {3}{2}\right)$.