Tag: mid-point formula

Questions Related to mid-point formula

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. $(0,0)$


Correct Option: D
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 $OP=\left(\dfrac{0+0}{2},\dfrac{4+-4}{2}\right)$
$=\left(0,0\right)$

Find the mid point of $(9,5)$ and $(3,7)$

  1. $(6,6)$

  2. $(12,12)$

  3. $(2,2)$

  4. $(1,1)$


Correct Option: A
Explanation:

Given points $(9,5),(3,7)$

Mid point is given as $\left(\dfrac{x _1+x _2}2,\dfrac{y _1,y _2}{2}\right)\\left(\dfrac{9+3}{2},\dfrac{5+7}{2}\right)\\left(\dfrac{12}{2},\dfrac{12}{2}\right)=(6,6)$

The mid point of $(8,3)$ and $(4,9)$ is 

  1. $(6,6)$

  2. $(4,4)$

  3. $(9,9)$

  4. $(2,2)$


Correct Option: A
Explanation:

The given points are $(8,3)$ and $(4,9)$

The mid point is given as$ \left(\dfrac {8+4}2,\dfrac {3+9}2\right)\(6,6)$

The mid point of $(-1,-3)$ and $(3,7)$

  1. (1, 2) 

  2. (0, 2) 

  3. (0, 4)

  4. (2 ,2)


Correct Option: A
Explanation:

Given points $(-1,-3),(3,7)$

Mid point is given as $\left(\dfrac {-1+3}{2},\dfrac {-3+7}{2}\right)=(1,2)$

The mid point of $(4,9)  $ and $(8,3)$ is 

  1. $(6,6)$

  2. $(5,7)$

  3. $(-6,-6)$

  4. None.


Correct Option: A
Explanation:

The mid point of $(4,9)  $ and $(8,3)$ is given as 

$\left(\dfrac{4+8}2,\dfrac{9+3}2\right)\\left(\dfrac {12}2,\dfrac{12}2\right)=(6,6)$

The mid point of $(2,3)$ and $(8,9)$ is 

  1. $(5,6)$

  2. $(2,8)$

  3. $(5,7)$

  4. $(4,6)$


Correct Option: A
Explanation:

The points are $(x _1,y _1)=(2,3)$ and $(x _2,y _2)=(8,9)$


The mid point is given as $\left(\dfrac{2+8}2,\dfrac {3+9}2\right)$
                                
$\left(\dfrac {10}2,\dfrac {12}2\right)=(5,6)$

The mid point of $(3,4)$ and $(1,-2)$

  1. (2,1)

  2. (1,2)

  3. (2,-1)

  4. (1,-2)


Correct Option: A
Explanation:

The points are $(3,4)$ and $(1,-2)$


The mid point of $(3,4)$ and $(1,-2)$ is given by 

$\left(\dfrac {x _1+x _2}2,\dfrac {y _1+y _2}2\right)\\\left(\dfrac{3+1}2,\dfrac {4-2}2\right)=(2,1)$

The mid-point of the line segment joining $( 2a, 4)$ and $(-2, 2b)$ is $(1, 2a + 1 )$. The values of $a$ and $b$ are 

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

  2. $a = 2, b = -3$

  3. $a = 3, b = - 2$

  4. $a =2, b = 3$


Correct Option: D
Explanation:

Midpoint of two points $ =\left( \cfrac { { x } _{ 1 }+{ x} _{ 2 } }{ 2 } ,\cfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Given, midpoint of $ (2a,4) $ and $ (-2,2b) = (1,2a+1) $
$ => \left(\cfrac { 2a-2 }{ 2 } ,\cfrac { 4+2b }{ 2 }\right)= (1,2a+1) $
$ => \cfrac { 2a-2 }{ 2 } = 1 ; \cfrac { 4+2b }{ 2 } = 2a + 1 $
$ => 2a -2 = 2 $
$=> a = 2 $

And, $ \cfrac { 4+2b }{ 2 } = 2a + 1 $
$=> \cfrac { 4+2b }{ 2 } = 2(2) + 1 = 5 $
$ => 4 + 2b = 10 $
$ => 2b = 6 $
$=> b = 3 $

The point which lies in the perpendicular bisector of the line segment joining the points A (-2, -5)  and B (2,5) is 

  1. (0, 0)

  2. (0, 2)

  3. (2, 0)

  4. (-2, 0)


Correct Option: A
Explanation:

A perpendicular bisector of a line segment, passed through its midpoint.

If C is the point on AB, through which its perpendicular bisector passes, then C $ = $ mid point of AB.

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 AB $= \left( \frac { -2+2 }{ 2 } ,\frac { -5+5 }{ 2 }  \right)

\quad =\quad (0,0) $



If Q$\displaystyle \left ( \frac{a}{3},4 \right )$ is the mid-point of the line segment joining the points A(-6,5) and B(-2,3), then the value of 'a' is

  1. 4

  2. -6

  3. -8

  4. -12


Correct Option: D
Explanation:

The co-ordinates of the mid-point  of the line segment joining the point $P(x _1,y _1),Q(x _2,y _2)$$=\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2}\right)$

$\Rightarrow \left(\dfrac{a}{3},4\right)=\left(\dfrac{-6+(-2)}{2},\dfrac{5+3}{2}\right)$
$\Rightarrow \left(\dfrac{a}{3},4\right)=\left(\dfrac{-8}{2},4\right)$
X co-ordinates
$\Rightarrow \dfrac{a}{3}=\dfrac{-8}{2}$
$\Rightarrow a=\dfrac{-8\times 3}{2}=-12$