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The line segment joining $A( 6,\ 3)$ and $B( -1,\ -4)$ is doubled in length by adding half of $AB$ to each end. Find the coordinates of the new end points.
Given: The line segment joining $A( 6,\ 3)$ and $B( -1,\ -4)$ is doubled in length by adding half of $AB$ to each end.
To do: To find the coordinates of the new end points.
Solution:
$AB:AC=1:2=m:n$.
$A( 6,\ 3)=( \frac{2x1-1}{2+1},\ \frac{2y1-4}{2+1})$
$\Rightarrow ( 6,\ 3)=( \frac{2x_1-1}{3},\ \frac{2y_1-4}{3})$
equating on both side,
$\Rightarrow \frac{2x_1-1}{3}=6;\ \frac{2y_1-4}{3}=3$
$\Rightarrow 2x_1-1=18;\ 2y_1-4=9$
$\Rightarrow 2x_1=18+1;\ 2y_1-4=9+4$
$\Rightarrow 2x_1=19;\ 2y_1=13$
$\Rightarrow x_1=\frac{19}{2};\ y_1=\frac{13}{2}$
$AB:BD=2:1$
by formula
$( -1,\ -4)=( \frac{2x_2+6}{2+1},\ \frac{2y_2+3}{2+1}]$
$( -1,\ -4)=( \frac{2x_2+6}{3},\ \frac{2y_2+3}{3})$
equating on both sides.
$\Rightarrow \frac{2x_2+6}{3}=-1;\ \frac{2y_2+3}{3}=-4$
$\Rightarrow 2x_2+6=-3;\ 2y_2+3=-12$
$\Rightarrow 2x_2=-3-6;\ 2y_2=-12-3$
$\Rightarrow 2x_2=-9;\ 2y_2=-15$
$x_2=-\frac{9}{2};\ y_2=-\frac{15}{2}$
therefore $C=( \frac{19}{2},\ \frac{13}{2})$
$D=(-\frac{9}{2},\ -\frac{15}{2})$
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