Question

find the point that divides AD where A(5,2) B(3,4) in the ratio 1:6 internally​

Answer 1

SOLUTION:-

The coordinates of the point P(x,y) which divides the line segment AB internally in the ratio m:n given by;

$= > (\frac{ mx2 + nx1}{m + n} , \frac{my2 + ny1}{m + n} )$

Let point C=C(x,y) divide the given line AB in the ratio 1:6.

AC:CB =m:n = 1:6.

Therefore,

$x = \frac{mx2 + nx1}{m + n} \\ \\ = > \frac{1 \times3 + 6 \times 5 }{1 + 6} \\ \\ = > \frac{3 + 30}{7} \\ \\ = > \frac{33}{7} = 4.7$

And y coordinate;

$y = \frac{my2 + ny1}{ m + n} \\ \\ = > \frac{1 \times4 + 6 \times 2 }{1 + 6} \\ \\ = > \frac{4 + 12}{7} \\ \\ = > \frac{16}{7} = 2.2$

Hence,

The coordinates of the point C are

(4.7, 4.8).

Hope it helps ☺️