Question

Oint A is located at (2, 6) and point B is located at (18, 12).

Answer 1

Solution:

Coordinates of point A = (2,6)

Coordinates of Point B = (18,12)

Let the point P(x,y) divides AB in the ratio of 2:3.

So, formula for internal division

x = $\frac{mx_{2}+nx_{1}}{m+n}$

y =  $\frac{my_{2} +ny_{1}}{m+n}$

x=

$\frac{3\times2+2\times18}{2+3}=\frac{6+36}{5}=\frac{42}{5}\\\\y=\frac{3 \times 6 + 2 \times 12}{2+3}=\frac{18+24}{5}=\frac{42}{5}$

The coordinates of point which divides AB in the ratio of 2:3 is $(\frac{42}{5}, \frac{42}{5})$.

Answer 2

Answer:

$(\frac{42}{5},\frac{42}{5})$

Step-by-step explanation:

We are given that

Point A is at (2,6) and  point B is at  (18,12).

We have to find the point which partitions the directed line segment AB into a ratio 2:3.

We have $x_1=2, y_1=6,x_2=18,y_2=12,m_1=2,m_2=3$

Section formula:$x=\frac{m_1x_1+m_2x_2}{m_1+m_2}, y=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Substitute the values in the given formula then, we get

The coordinates of point which partitions the segment AB is given by

$x=\frac{2(18)+3(2)}{2+3}, y=\frac{2(12)+3(6)}{2+3}$

$x=\frac{36+6}{5},y=\frac{24+18}{5}$

$x=\frac{42}{5}, y=\frac{42}{5}$

Hence, the coordinated of the point which partitions the line segment AB =$(\frac{42}{5},\frac{42}{5})$