Question

A(2, 4) and B(8, 12) are two ends of a line segment. Find the point which divides AB internally in the ratio 1 : 3.
$\left[3\right]$

Answer 1

Solution:
Let the point $P(x,y)$ divide AB in the ratio 1 : 3.

The coordinates of the point that divides a line segment internally , joining the points $(x_1,y_1)and(x_2,y_2)$ in the ratio m : n are:

$(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n})$
[1 Mark]

Here,
$x_1=2,y_1=4x_2=8,y_2=12m=1,n=3$
[0.5 Marks]

So, $(x,y)=(\frac{1\times 8+3\times 2}{1+3},\frac{1\times 12+3\times 4}{1+3})$
[0.5 Marks]
$=(\frac{14}{4},\frac{24}{4})=(\frac{7}{2},6)$

$\therefore$ The point that divides AB in the ratio 1 : 3 is $(\frac{7}{2},6)$.
[1 Mark]