Question

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.

$\left[5\right]$

Answer 1

Let the end points be $P(x_1,x_2),Q(x_2,y_2)$

Let AB=2k then PA=BQ=k
$\Rightarrow \frac{PA}{AB}=\frac{k}{2k}=\frac{1}{2}$ and $\frac{AB}{BQ}=\frac{2k}{k}=\frac{2}{1}$
$\left[1\right]$

A(6,3) is dividing line segment PB internally in the ratio 1 : 2.

Using section formula for internal divison i.e.,

$(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$\Rightarrow (6,3)=\left(\frac{\left(1\right)(-1)+\left(2\right)\left(x_1\right)}{1+2},\frac{\left(1\right)(-4)+\left(2\right)\left(y_1\right)}{1+2}\right)\Rightarrow (6,3)=\left(\frac{-1+2x_1}{3},\frac{-4+2y_1}{3}\right)\Rightarrow 6=\frac{-1+2x_1}{3}and3=\frac{-4+2y_1}{3}\Rightarrow 2x_1=19and2y_1=13\Rightarrow x_1=\frac{19}{2}and{y}_1=\frac{13}{2}$

Hence, the coordinates of end point P are $\left(\frac{19}{2},\frac{13}{2}\right)$
$\left[2\right]$

B(-1,-4) is dividing line segment AQ internally in the ratio 2 : 1.

Again by using section formula, we can write as

$\Rightarrow (-1,-4)=\left(\frac{\left(2\right)\left(x_2\right)+\left(1\right)\left(6\right)}{2+1},\frac{\left(2\right)\left(y_2\right)+\left(1\right)\left(3\right)}{2+1}\right)\Rightarrow (-1,-4)=\left(\frac{2x_2+6}{3},\frac{2y_2+3}{3}\right)\Rightarrow -1=\frac{2x_2+6}{3}and-4=\frac{2y_2+3}{3}\Rightarrow 2x_2=-9and2y_2=-15\Rightarrow x_2=\frac{-9}{2}and{y}_2=\frac{-15}{2}$

Hence, the coordinates of another end point Q are $\left(\frac{-9}{2},\frac{-15}{2}\right)$
$\left[2\right]$