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Mid-point of a line segment - class - XI

Description: mid-point of a line segment
Number of Questions: 22
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Tags: transformations straight lines and quadratic equations three dimensional coordinates coordinate geometry three dimensional geometry straight lines maths introduction to three dimensional geometry introduction to three-dimensional geometry geometry and algebra section and mid-point formula coordinates points and lines
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The ordinate of the point which divides the lines joining the origin and the point $(1,2)  $ externally in the ratio of $3:2$ is

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $-2$

  2. $ \displaystyle \frac{3}{5} $

  3. $ \displaystyle \frac{2}{5} $

  4. $6$

Show answer
Correct Option: D
Explanation:

Co-ordinates of the required point will be
$\displaystyle y=\frac{m _{1}y _{2}-m _{2}y _{1}}{m _{1}-m _{2}}=\frac{3\times 2-2\times 0}{3-2}=6$

The points (22,23) divides the join of P (7,5) and Q externally in the ratio 3:5, then Q=

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(3,7)$

  2. $(-3,7)$

  3. $(3,-7)$

  4. $(-3,-7)$

Show answer
Correct Option: D
Explanation:
$\left(22,23\right)$dives $P\left(7,5\right)$ and $Q$ in the ratio $3:5$ externally.
Let $Q\left(x,y\right)$
$22=\dfrac{3x-5\times 7}{-2}$
$-44=3x-35$
$x=-3$
$23=\dfrac{3y-5\times 5}{-2}$
$y=-7$
$Q\left(-3,-7\right)$
$D$ is correct.

The point $(22, 33)$ divides the join of $P(7, 5)$ and $Q$ externally in the ratio $3 : 5$, then coordinates of $Q$ are

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(3, 7)$

  2. $( - 3, -7)$

  3. $(3, - 7)$

  4. $( - 3, \frac{ - 41}{3})$

Show answer
Correct Option: B
Explanation:
Let the point $Q$ be $\left(x,y \right)$
As the point $\left(22,23 \right)$ diverts the line joining point $P\left(7,5 \right)$ and $Q \left(x,y \right)$ in $3:5$ externally.
So.  
$22=\dfrac{ 3\left(x\right)-5\left(7\right)} {3-5}\quad 23=\dfrac{ 3\left(y\right)-5\left(5\right)} {3-5}$
$22=\dfrac{3x-35}{-2}\quad 23=\dfrac{3y-25}{-2}$
$-44=3x-35\quad -46=3y-25$
$-=3x\quad -21=3y$
$x=-3\quad y=\dfrac{-21}{3}=-7$
Point $Q=\left(-3,-7\right)$

Find the co-ordinates of the point $P$ which divides segment $JL$ externally in the ratio $m:n$ in the following example:

$J(5, -3), L(0, 9), m:n = 4:3$

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(-15, 45)$

  2. $(15, -45)$

  3. $(15, 45)$

  4. $(-15, -45)$

Show answer
Correct Option: A
Explanation:
Let the co-ordinates of the point $P$ be $P\left( x,y \right)$  which divides the line segment $JL$ joining the points $J\left( { x } _{ 1 },{ y } _{ 1 } \right) =\left( 5,-3 \right)  \&  L\left( { x } _{ 2 },{ y } _{ 2 } \right) =\left( 0,9 \right)$ in the ratio $m:n=4:3$

Then, by the section formula, $ x=\dfrac { n{ x } _{ 1 }-m{ x } _{ 2 } }{ n-m } =\dfrac { 3\times 5-4\times 0 }{ 3-4 } =-15$ and 
$ y=\dfrac { n{ y } _{ 1 }-m{ y } _{ 2 } }{ n-m } =\dfrac { 3\times (-3)-4\times 9 }{ 3-4 } =45$ 
$\therefore  P\left( x,y \right) =P\left( -15,45 \right)  $

The co-ordinates of the point B which divides segment PQ joining the points $P(-2,-4)$ and $Q(-2,-1)$ externally in the ratio $m: n=7:1$ are

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $ B(x,y)=\left( 2,-\dfrac { 1 }{ 2 } \right) $

  2. $ B(x,y)=\left( -2,\dfrac { 1 }{ 2 }\right) $

  3. $ B(x,y)=\left( -2,-\dfrac { 1 }{ 2 } \right) $

  4. $ B(x,y)=\left( 2,\dfrac { 1 }{ 2 }\right)$

Show answer
Correct Option: C
Explanation:

Given that- the point $B\left( x,y \right)$ externally divides the line segment PQ joining the points $P(-2,-4)$ & $Q\left( { x } _{ 2 },{ y } _{ 2 } \right) =(-2,-1)$ in the ratio $m:n=7:1$

To find out- the coordinates of B.
Solution-  We know that if a point $B\left( x,y \right)$ externally divides the line segment PQ joining the points $P\left( { x } _{ 1 },{ y } _{ 1 } \right)$ & 
$Q\left( { x } _{ 2 },{ y } _{ 2 } \right)$ in the ratio $m:n$ then, by the section formula, 
$x=\dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n }$ &  $y=\dfrac { m{ y } _{ 2 }-n{ y } _{ 1 } }{ m-n }$
Here ${ x } _{ 1 }=-2, { x } _{ 2 }=-2, { y } _{ 1 }=-4, { y } _{ 2 }=-1, m=7, n=1$
$ \therefore  x=\dfrac { 7\times \left( -2 \right) -1\times \left( -2 \right)  }{ 7-1 } =-2$ 
And $y=\dfrac { 7\times (-1)-1\times \left( -4 \right)  }{ 7-1 } =-\dfrac { 1 }{ 2 }$
$\therefore  B(x,y)=\left( -2,-\dfrac { 1 }{ 2 }  \right)$

If $A(-2,5)$ and $B(3,2)$ are the points on a straight line. If ${AB}$ is extended to $'C'$ such that $AC=2BC$, then the co-ordinates of $'C'$ are ____

maths transformations midpoint of line segment mid point formula mid-point of a line segment

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

  2. $\left(\displaystyle\frac{7}{2}, \frac{1}{2}\right)$

  3. $(8, -1)$

  4. $(-1, 8)$

Show answer
Correct Option: C
Explanation:

Given points are $A(-2,5)$ and $B(3,2)$ 
$\dfrac{AC}{BC}=\dfrac{2}{1}$ , $C$ divides line segment $ { AB } $ externally.
If $A({ x } _{ 1 },{ y } _{ 1 })$ and $B({ x } _{ 2 },{ y } _{ 2 })$ be two end points of a line segment, then the coordinates of the point $P(x,y)$ that divides the line segment externally in the ratio $m:n$ is $\left( \dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n } ,\dfrac { m{ y } _{ 2 }-ny _{ 1 } }{ m-n }  \right) $.
Thus, the coordinates of $C$ are $\left( \dfrac { 2(3)-1(-2) }{ 2-1 } ,\dfrac { 2(2)-1(5) }{ 2-1 }  \right) $.
$=\left( \dfrac { 6+2 }{ 1 } ,\dfrac { 4-5 }{ 1 }  \right) $
$=(8,-1)$

The co-ordinates of the point B which divides segment PQ joining the points $P(-2,-4)$ and $Q(-2,-1)$ in the ratio $m:n = 2 : 5$,  are

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $B(x,y)=(2, 6)$

  2. $B(x,y)=(-2,6)$

  3. $B(x,y)=(2,-6)$

  4. $B(x,y)=(-2,-6)$

Show answer
Correct Option: D
Explanation:

Given that:

Point $B\left( x,y \right)$  externally divides the line segment $PQ$ joining the points $P(-2,-4)$ &  $Q\left( { x } _{ 2 },{ y } _{ 2 } \right) =(-2,-1)$ in the ratio $m:n=2:5$. 

To find out- the co-ordinates of B.

Solution-
We know that if a point $B\left( x,y \right)$ externally divides the line segment PQ joining the points $P\left( { x } _{ 1 },{ y } _{ 1 } \right)$ & $Q\left( { x } _{ 2 },{ y } _{ 2 } \right)$ in the ratio $m:n$ then, by the section formula, 
$x=\dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n }$ & $y=\dfrac {my _2-ny _1}{m-n}$

Here, ${ x } _{ 1 }=-2, { x } _{ 2 }=-2, { y } _{ 1 }=-4, { y } _{ 2 }=-1, m=2, n=5.$

$\therefore  x=\dfrac { 2\times (-2)-5\times (-2) }{ 2-5 } =-2$ and $y=\dfrac { 2\times (-1)-5\times (-4) }{ 2-5 } =-6$

$\therefore  B(x,y)=\left( -2,-6\right)$

Find the co-ordinates of the point dividing the join of $A(1, -2)$ and $B(4, 7)$ externally in the ratio of $2 : 1.$

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(7, 16)$

  2. $(7,12)$

  3. $\left(3,\displaystyle \frac{16}{3}\right)$

  4. $(3,16)$

Show answer
Correct Option: A
Explanation:
The given points are $A(1,-2)$ and $B(4,7)$. 
We have to find the coordinate of points which divide the line segment externally in the ratio $2:1$
Let point $C$ divides the segment $AB$ in the ratio $1:2$, hence $m=2, n=1$
By section formula which states that when the line segment is divided externally by the point in the ration $m:n$ then coordinates of point are

$\Rightarrow$  $C=\left(\dfrac{mx _2-nx _1}{m-n},\,\dfrac{my _2-n _1}{m-n}\right)$

           $=\left(\dfrac{(2)(4)-(1)(1)}{2-1},\,\dfrac{(2)(7)-(1)(-2)}{2-1}\right)$

           $=\left(\dfrac{7}{1},\dfrac{16}{1}\right)$

           $=(7,16)$

The point (11, 10) divides the line segment joining the points (5, -2) and (9, 6) in the ratio

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. 1 : 3 internally

  2. 1 : 3 externally

  3. 3 : 1 internally

  4. 3 : 1 externally

Show answer
Correct Option: D
Explanation:

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x } _{ 1 },{ y } _{ 1

})$ and $({ x } _{ 2 },{ y } _{ 2 })$ in the ratio $ m:n $, then $(x,y) =

\left( \dfrac { m{ x } _{ 2 } + n{ x } _{ 1 } }{ m + n } ,\dfrac { m{ y } _{ 2

}  + n{ y } _{ 1 } }{ m + n }  \right) $

Let the ratio be $ k : 1 $


Substituting $({ x } _{ 1 },{ y } _{

1 }) = (5,-2) $ and $({x } _{ 2 },{ y } _{ 2 }) = (9,6) $  in the

section formula, we get  $ \left( \dfrac { k(9)  + 1(5) }{ k + 1 }

,\dfrac { k(6) + 1(-2) }{ k + 1 }  \right) = ( 11,10) $ 


$ \left( \dfrac { 9k + 5}{ k + 1 }

,\dfrac { 6k - 2 }{ k + 1} \right) = ( 11,10) $


Comparing the x - coordinate,

$ => \dfrac { 9k + 5 }{ k + 1 } = 11 $

$ =>9k + 5 = 11k + 11 $


$ 2k = -6 $


$ k = -3 $

Hence, the ratio is $ 3:1$ externally.

Value of m for which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8) is

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. 5

  2. $\displaystyle \frac{3}{2}$

  3. $\displaystyle -\frac{2}{5}$

  4. None

Show answer
Correct Option: C
Explanation:

Equation

of a line joining two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2

},{ y } _{ 2 }) $ is given by the formula $ y-{ y } _{ 1 }=\quad \left( \dfrac {

{ y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-{ x } _{ 1 } }  \right) (x-{ x } _{ 1

}) $

Equation of line passing through A $(-4,3)

$ and B $ (2,8)$  is  $ y-3=\quad \left( \dfrac { 8-3 }{ 2+4 }  \right) (x+4) $
$

=> y-3=\dfrac { 5 }{ 6 } (x+4) $

$ => 6y-18=5x+20$

$ => 5x-6y+38=0$

Since Point $  P(m, 6)  $ divides this line, it should satisfy the equation of the line, if we substitute

$ x = m $ and $ y =6 $ in it.


So, $  5m-36+38=0 $

$ => 5m = -2 $

$ m = -\dfrac {2}{5} $

Find the coordinates of the point which divides the line segment joining the points (6, 3) and (-4, 5) in the ratio 3 : 2 externally.
maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. (-12, 9)

  2. (-16, 9)

  3. (-24, 9)

  4. (-14, 9)

Show answer
Correct Option: C
Explanation:

Using the section formula, if a point $(x,y)$ divides the line joining the points

$({ x } _{ 1 },{ y } _{ 1 })$ and $({ x } _{ 2 },{ y } _{ 2 })$externally  in the ratio $ m:n $, then $(x,y) = \left(

\dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n } ,\dfrac { m{ y } _{ 2 }-n{ y } _{ 1 }

}{ m-n }  \right) $


Substituting $({ x } _{ 1 },{ y } _{ 1 }) = (6,3) $ and

$({x } _{ 2 },{ y } _{ 2 }) = (-4,5) $  and $ m = 3, n = 2 $ in the section formula, we get 



$ C = \left( \dfrac { 3(-4)-2(6) }{ 3-2 } ,\dfrac { 3(5)-2(3) }{ 3-2 } 

\right) =\left( -24,9 \right) $

If the line joining A(2, 3) and B(-5, 7) is cut by x-axis at P then AP : PB is

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. 3 : 7

  2. -3 : 7

  3. 7 : 3

  4. 7 : -3

Show answer
Correct Option: B
Explanation:

Using the section formula, if a

point $(x,y)$ divides the line joining the points $({ x } _{ 1 },{ y } _{ 1

})$ and $({ x } _{ 2 },{ y } _{ 2 })$ in the ratio $ m:n $, then $(x,y) =

\left( \dfrac { m{ x } _{ 2 } + n{ x } _{ 1 } }{ m + n } ,\dfrac { m{ y } _{ 2

}  + n{ y } _{ 1 } }{ m + n }  \right) $


Substituting $({ x } _{ 1 },{ y } _{ 1 }) = (2,3) $ and $({x } _{ 2 },{ y } _{ 2

}) = (-5,7) $  in the section formula, we get the point $ \left( \dfrac {

m(-5)  + n(2) }{ m + n } ,\dfrac { m(7) + n(3) }{ m + n }  \right)

=\left( \dfrac { -5m  + 2n }{ m + n } ,\dfrac { 7m + 3n }{ m + n} \right) $


As the point lies on x - axis, y -coordinate $ = 0 $.

$ => \dfrac { 7m + 3n }{ m + n} = 0 $ 

$ => 7m = -3n $  or $ m : n = -3:7 $

Find the coordinates of the point which divides the line segment joining the points $(6, 3)$ and $(-4, 5)$ in the ratio $3 : 2$, externally.

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(24,9)$

  2. $(-24,-9)$

  3. $(-24,9)$

  4. $(24,-9)$

Show answer
Correct Option: C
Explanation:

Let P$(x,y)$ be the required point.
Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x } _{ 1 },{ y } _{ 1 })$ and $({ x } _{ 2 },{ y } _{ 2 })$externally  in the ratio $ m:n $, then $(x,y) = \left( \dfrac { m{ x } _{ 2 }-n{ x } _{ 1 } }{ m-n } ,\dfrac { m{ y } _{ 2 }-n{ y } _{ 1 } }{ m-n }  \right) $
Substituting $({ x } _{ 1 },{ y } _{ 1 }) = (6,3) $ and $({x } _{ 2 },{ y } _{ 2 }) = (-4,5) $  and $ m = 3, n = 2 $ in the section formula, we get 

$ P = \left( \dfrac { 3(-4)-2(6) }{ 3-2 } ,\dfrac { 3(5)-2(3) }{ 3-2 } \right) =\left(-24,9 \right) $

The ordinate of the point which divides the line joining the origin and the point (1, 2) externally in the ratio of 3 : 2 is

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $-2$

  2. $\displaystyle\frac{3}{5}$

  3. $\displaystyle\frac{2}{5}$

  4. $6$

Show answer
Correct Option: D
Explanation:

The co-ordinates of the required point will be

$\displaystyle y=\dfrac{m _1y _2-m _2y _1}{m _1-m _2}$

$\displaystyle=\dfrac{3\times2-2\times0}{3-2}=6$

Find the co-ordinates of a point C on AB produced such that $3AB = AC$, where $A = (3, 2)$ and $B = (-2, 4).$

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(-12, 8)$

  2. $(8, 12)$

  3. $(12, 8)$

  4. $(-8, 12)$

Show answer
Correct Option: A
Explanation:

From the above condition, we can observe that the point B  divides the line segment joining AC in 1;3 ratio, or $AC:AB=3:1$ or $AB:BC=2:1$. Let the coordinates of C be (x,y). Therefore,
$B(-2,4)=\left(\dfrac{1(x)+2(3)}{3},\dfrac{1(y)+2(2)}{3}\right)$
Or  $\dfrac{6+x}{3}=-2$ or $6+x=-6$ or $x=-12$. Similarly $\dfrac{4+y}{3}=4$ or $4+y=12$ or $y=8$.
Hence $C(x,y)=(-12,8)$

Find $x$ and $y$ if $(2,5)$ is the midpoint of points $(x,y)$ and $(-5,6)$.

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $x=4, y=9$

  2. $x=9, y=4$

  3. $x=-9, y=4$

  4. $x=9, y=-4$

Show answer
Correct Option: B
Explanation:
If the end points of a line segment is $(x,y)$ and $(-5,6)$ then the midpoint of the line segment has the coordinates:
$\left( \dfrac { x-5 }{ 2 } ,\dfrac { y+6 }{ 2 }  \right) =\left( 2,5 \right)$ ...(hint: using mid-point formula)
Now equating the points:  $\dfrac { x-5 }{ 2 } =2$ 
$\Rightarrow x-5=4$ 
$\Rightarrow x=9$
And, $\dfrac { y+6 }{ 2 } =5$ 
$\Rightarrow y+6=10$ 
$\Rightarrow y=4$

Hence, $x=9$ and $y=4$.

Find the coordinates of the point which divides the join of the points $(2,4)$ and $(6,8)$ externally in the ratio $5:3$.

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. $(12,14)$

  2. $(14,12)$

  3. $(-12,14)$

  4. $(12,-14)$

Show answer
Correct Option: A
Explanation:

Given $A(2, 4)$ and $B(6,8)$

Applying the section formula externally,

$\left( \dfrac { L{ x } _{ 2 }-{ mx } _{ 1 } }{ L-m } ,\dfrac { L{ y } _{ 2 }-{ my } _{ 1 } }{ L-m }  \right)$

Here the ratio given is $5:3$ that is $L=5$ and $m=3$, therefore,

$\left( \dfrac { L{ x } _{ 2 }-{ mx } _{ 1 } }{ L-m } ,\dfrac { L{ y } _{ 2 }-{ my } _{ 1 } }{ L-m }  \right) =\left( \dfrac { (5\times 6)-(3\times 2) }{ 5-3 } ,\dfrac { (5\times 8)-(3\times 4) }{ 5-3 }  \right) $


$=\left( \dfrac { 30-6 }{ 2 } ,\dfrac { 40-12 }{ 2 }  \right) =\left( \dfrac { 24 }{ 2 } ,\dfrac { 28 }{ 2 }  \right) =\left( 12,14 \right)$ 

Hence, the coordinates of the point is $(12,14)$.

If the join of the two points $(x _1, y _1)$, $(x _2, y _2)$ is divided by a point R externally in ratio $m : n$ then

maths transformations midpoint of line segment mid point formula mid-point of a line segment

  1. x - coordinates is $\dfrac {mx _2 - nx _1}{m - n}$

  2. x - coordinates is $\dfrac {my _2 - ny _1}{m - n}$

  3. Both (a) and (b) above

  4. None of these

Show answer
Correct Option: A
Explanation:

When a point C divides a segment $A(x _1,y _1)$ and $B(x _2,y _2)$ in the ratio $m:n$ externally, we use the section formula to find the coordinates of that point.

The Coordinates of point R will be,

$X=\dfrac{mx _2-nx _1}{m-n}$.

If $z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$, then

maths introduction to three dimensional geometry midpoint of line segment mid point formula mid-point of a line segment

  1. $\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$

  2. $\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$

  3. $\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$

  4. $\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$

Show answer
Correct Option: B
Explanation:

if $z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)$ then.

As $z=|z|e^{i arq (z)}$
$\therefore z=\cos \left(\dfrac{\pi}{6}\right)+i\sin \left(\dfrac{\pi}{6}\right)=e^{i\left(\pi/6\right)}\quad [\because e^{i\theta}=\cos\theta+i\sin \theta]$
$\Rightarrow |z|=1,  arq=\dfrac{\pi}{6}$

STATEMENT - 1 : The coordinates of the point P(x, y) which divides the line segment joining the points A$(x _1,  y _1)$ and B$(x _2,  y _2)$ internally in the ration $m _1$  :  $m _2$ are $\left ( \dfrac{m _1 x _2 -m _2 x _1}{m _1 + m _2} ,  \dfrac{m _1 y _2 - m _2 y _1}{m _1 + m _2}\right )$


STATEMENT - 2 : The mid-point of the line segment joining the points P $(p _1 y _1)$ and Q$(x _2, y _2)$ is $\left ( \dfrac{x _1+x _2}{2} , \dfrac{y _1 + y _2}{2} \right )$

maths section and mid-point formula midpoint of line segment mid point formula mid-point of a line segment

  1. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

  2. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

  3. Statement - 1 is True, Statement - 2 is False

  4. Statement - 1 is False, Statement - 2 is True

Show answer
Correct Option: D
Explanation:

Statement -1 is false,

As the formula is not for the internally it is when point divides externally.
Statement -2 is true.

The ratio in which the joining of (-3,2) and (5,6) is divided by the y-axis is

maths section and mid-point formula midpoint of line segment mid point formula mid-point of a line segment

  1. 3:5

  2. 2:5

  3. 1:3

  4. 2:3

Show answer
Correct Option: A
Explanation:

Let the ratio be k:1 Then
x-coordinate of P = 0

$\dfrac{k\times 5+1\times -3}{k+1}=0$
$k=\dfrac{3}{5}$

Consider points $A(-1,3), B(-1,2)$. Find point $P$ which divides $AB$ externally in $\dfrac{5}{4}$.

maths section and mid-point formula midpoint of line segment mid point formula mid-point of a line segment

  1. $(9,-22)$

  2. $(-1,2)$

  3. $(-1,-2)$

  4. $(9,22)$

Show answer
Correct Option: C
Explanation:
Let point P be (x,y)
$x=\cfrac { 5\times (-1)-4\times (-1) }{ 5-4 } \\ x=\cfrac { -5+4 }{ 1 } \\ x=-1\\ y=\cfrac { 5\times (2)-4\times (3) }{ 5-4 } \\ y=\cfrac { 10-12 }{ 1 } \\ y=-2\\ \therefore P=(-1,-2)$
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