Question

18. The points of trisection of A(0,1), B(0,4) are​

Answer 1

Let the points be = P(x, y) and Q(x', y')

Refer to the attachment for the figure.

We notice that AP:PB = 1:2

→ By using the section formula:-

$\sf{ (x,y) = (\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}) , (\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}})}$

→ $\boxed{\sf{(x,y) = (0,2)}}$

Also, PQ:QB = 1:1

→ By using the mid point formula:-

$\sf { (x,y) = (\frac{x_{1} +x_{2}}{2}), (\frac{y_{1}+y_{2}}{2}) }$

→ $\boxed{\sf{(x',y') = (0,3)}}$

Thus, the points are (0,2) and (0,3)