Tag: transformations

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.