Question

Find the sum of co-ordinates of the point which divides the join of the points (2, 4) and (6, 8) externally in the ratio 5 : 3.

Answer 1

The required coordinates of the point which divides the join (2, 4) and (6, 8) externally in the ratio 5 : 3, are
$(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n})$
Here $m:n=5:3,(x_1,y_1)=(2,4)(x_2,y_2)=(6,8)$
Hence the required co-ordinates
$=(12,14)$