Tag: mid-point of a line segment

Questions Related to mid-point of a line segment

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} $