Tag: transformations

Questions Related to transformations

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) $