Section Formula and Midpoint of a Line Segment - Class XI

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

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 $m:n=7:1$

Given that:

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.

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} $

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 

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 $

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 co-ordinates of the required point will be

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

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

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 $C(x,y)=(-12,8)$

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 $(12,14)$.

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.

Statement -1 is false,

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