Midpoints - class-XI

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

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

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

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

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

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

The points are $(3,4)$ and $(1,-2)$

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) $

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)$

Midpoint of BC = L = (4, 2)
$\displaystyle \therefore AL=\sqrt{\left ( 1-4 \right )^{2}+\left ( -1-2 \right )^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

Given,

Since, $(-2, -4)$ is the midpoint of $(6, -7)$ and $(x,y).$
$\Rightarrow \dfrac{x+6}2=-2\Rightarrow x=-10$
and $ \dfrac{y-7}2=-4\Rightarrow y=-1$
Option D is correct.

Given line segment point $(5,9)$ and $(7,11)$, then mid point as per section formula: 

Mid-point of $PR=\left(\cfrac{-2+3}2,\cfrac{2-2}2\right)=\left(\cfrac12,0\right)$

Mid point $=$ $\dfrac{x _1+x _2}{2}, \dfrac{y _1+y _2}{2}$

Mid point of points $(a+b,a-b),(-a,b)$ is $\left( \dfrac { a+b-a }{ 2 } ,\dfrac { a-b+b }{ 2 }  \right) =\left(\dfrac { b }{ 2 } ,\dfrac { a }{ 2 } \right)$

If $C$ is the midpoint of $AB$, then $C$ divides $AB$ in equal segments. Those segments are $AC$ and $AB$.

$O(0,0) $ and $P(-8,0)$ are given points.

Given points $A(2,3)\equiv(x _1,y _1)$ and $B(5,6)\equiv(x _2,y _2)$

$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x-1(y _2-y _3)+x _2(y _3+y _1)+x _3(y _1-y _2)]\\=2[0+2(3-1)+0]=8sq\>unit$

Rectangle's midpoints are $ = \left( {\frac{{0 + 4}}{2},\frac{{0 + 2}}{2}} \right) = \left( {\frac{4}{2},\frac{2}{2}} \right) = \left( {2,1} \right)$

Mid-point of $(a,2)$ and $(3,6)$ is $(2,b)$

In a parallelogram, the diagonals bisect each other. Let the vertices be A(3, -4), B(x, y), C(-6, 5), and D(2, 1). The midpoint of diagonal AC is ((3-6)/2, (-4+5)/2) = (-1.5, 0.5). The midpoint of diagonal BD is ((x+2)/2, (y+1)/2). Setting these equal gives x+2 = -3 (x=-5) and y+1 = 1 (y=0).

The center of the circle x^2 + y^2 - 3x + 8y - 4 = 0 is (3/2, -4). Since the center is the midpoint of the diameter, if one end is (6, -3), let the other be (x, y). Then (6+x)/2 = 3/2 implies x = -3, and (-3+y)/2 = -4 implies y = -5.

Let the point be (x, 0). Equidistance means (x-2)^2 + (0-3)^2 = (x-1)^2 + (0-5)^2. Expanding: x^2 - 4x + 4 + 9 = x^2 - 2x + 1 + 25. Simplifying: -4x + 13 = -2x + 26, so -2x = 13, x = -13/2. The point is (-13/2, 0).

The locus of the midpoint of a segment whose endpoints move on axes while the lines are perpendicular results in a straight line relationship based on the fixed points P1 and P2.

$P=(1,0), Q=(h,k)$

For a parabola y^2 = 4ax, the chord subtending a right angle at the vertex has the equation y = mx + 2am. The midpoint (h, k) of this chord satisfies k = mh + 2am and the property that the chord is y = (2a/k)x - 4a^2/k. Substituting and simplifying leads to the locus y^2 = 2a(x - 4a), which is y^2 - 2ax + 8a^2 = 0.

Let the point be (0, y). Equidistance means the square of the distances to (6,5) and (-4,3) are equal: (0-6)^2 + (y-5)^2 = (0-(-4))^2 + (y-3)^2. This simplifies to 36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9, which results in 61 - 10y = 25 - 6y, or 4y = 36, so y = 9.

Let the point on x-axis equidistant from (3,5) and (4,2) be (x,0),

Given that $KR$ is a line segment joining the points $K=(2.5, -4.3)$ and $R=(-1.5, 2.7)$ 

Mid-point of the other vertices is also the mid-point of the given vertices and hence satisfies the given relation
So, $\cfrac { b+3 }{ 2 } =2\left( \cfrac { a+5 }{ 2 }  \right) +c$
$\Rightarrow 2a+2c-b+7=0$
Also, ${ c }^{ 2 }+2a-b=0$
$\Rightarrow { c }^{ 2 }-2c-7=0\Rightarrow c=1\pm 2\sqrt { 2 } $

The centre of the circle lies at the mid point of the diameter.
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) $
Using this formula, mid point of $ (3,y) $ and $ (x,2) = \left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 } \right) $
So,$\left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 }  \right) = (-6, 1.5) $
$ \Rightarrow  - 3 + x = -6 \times 2 $ and $ y - 2 = 1.5 \times 2 $
$\Rightarrow x = - 9 ; y = 5 $

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) $
Using this formula, mid - point of the line-segment joining the points $(5,4)$ and $(9,8)$ is: $= \left( \cfrac { -5 + 9 }{ 2 } ,\cfrac { 4- 8 }{ 2 }  \right)  = (2,-2) $

Let the fourth vertex D $ = (x,y) $
We know that the diagonals of a parallelogram bisect each other. So,the

midpoint of AC is same as the mid point of BD.

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) $

So, midpoint of $ AC = $ Mid point of $ BD $

$ => \left( \frac { 1+5 }{ 2 } ,\frac { -2+10 }{ 2 }  \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => \left( \frac { 6 }{ 2 } ,\frac { 8 }{ 2 }   \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => 3+x=6 ; 6 + y = 8 $

$ => x = 3 ; y = 2 $

Hence, $ D = (3,2) $

The centre of the circle lies at the mid point of the diameter.

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) $
Using this formula, mid point of the diameter $= \left( \dfrac { -3 + 5}{ 2 } ,\dfrac { 8 + 6 }{ 2 }  \right) = (1,7) $

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) $

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) $

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 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 $(\cfrac{x _1+x _2}{2},\cfrac{y _1+y _2}{2})$

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

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 $\Rightarrow\begin{pmatrix}\dfrac{5+9}{2},,\dfrac{3+8}{2}\end{pmatrix}$
$\Rightarrow\begin{pmatrix}7,,\dfrac{11}{2}\end{pmatrix}$

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

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

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

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 $\left(5, 2\right)$

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
$\left(\dfrac{k+1}{2}\right)$ $= 1$
$\Rightarrow k + 1 = 2$
$\Rightarrow k = 2 - 1$
$\Rightarrow k = 1$

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

Given two points $(x _{1},y _{1})$ and $(x _{2},y _{2})$, then the coordinates of the midpoint are $\left (\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}\right)$.
In the above case the points are $(-3,0)$ and $(7,4)$. 

Given points $A$ $(\sqrt{2},4)$, $B$ $(6,-\sqrt{3})$ and $C$ are col-linear if $B$ the mid point $AC$. 

We know that the diagonal of square is intersect equal at mid point. 

Given, coordinates $A (-3,-8)$ and the mid point of $AB$ is $(1,-5)$

The intersection point of diagonals of rectangle is midpoint of each diagonal.

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

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.

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

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

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

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:

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

As $A\equiv\left( 0,b \right) ,B\equiv\left( 0,0 \right) $ and $C\equiv\left( a,0 \right) $ 
Then $D\equiv\left( \cfrac { a }{ 2 } ,0 \right) ,E\equiv\left( \cfrac { a }{ 2 } ,\cfrac { b }{ 2 }  \right) $ and $F\equiv\left( 0,\cfrac { b }{ 2 }  \right) $
Using $a^{ 2 }+b^{ 2 }=20$, we have
${ \left( AD \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }=\cfrac { { a }^{ 2 } }{ 4 } +{ b }^{ 2 }\ Similary,, {(CF)}^2\ { \left( AD \right)  }^{ 2 }+{ \left( CF \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }+{ \left( 0-a \right)  }^{ 2 }+{ \left( \cfrac { b }{ 2 } -0 \right)  }^{ 2 }\ =\cfrac { 5\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ 4 } =25={ \left( BE \right)  }^{ 2 }$

Let the given vertices be $ A (1,-12)  B (5, -10)  C(a,b) ; D(x,y) $

We know that the diagonals of a parallelogram bisect each other. So,the midpoint of AC $ = (2,-5) $ and of BD $ = (2,-5) $.