Question

Find the point B on AC such that the ratio of AB to BC is 2:1

Answer 1

The position of point B is:

`(x_(3), y_(3)) = \left((2)/(3)\cdot x_(2) + (1)/(3) \cdot x_(1), (2)/(3)\cdot y_(2) + (1)/(3) \cdot y_(1) \right)`

Explanation:

Let be
`A = (x_(1), y_(1))`,
`B = (x_(3), y_(3))` and
`C = (x_(2), y_(2))`. The ratio is:

`((x_(3)-x_(1),y_(3)-y_(1)))/((x_(2)-x_(3),y_(2)-y_(3))) = 2`

After some algebraic handling:

`(x_(3)-x_(1), y_(3)-y_(1)) = 2 \cdot (x_(2)-x_(3),y_(2)-y_(3))`

`(x_(3)-x_(1), y_(3)-y_(1)) = (2\cdot x_(2) - 2\cdot x_(3), 2\cdot y_(2) - 2\cdot y_(3))`

`(3\cdot x_(3), 3\cdot y_(3)) = (2\cdot x_(2) + x_(1), 2\cdot y_(2) + y_(1))`

`3\cdot (x_(3), y_(3)) = (2\cdot x_(2) + x_(1), 2\cdot y_(2) + y_(1))`

`(x_(3),y_(3)) = (1)/(3)\cdot (2\cdot x_(2)+x_(1), 2\cdot y_(2)+y_(1))`

The position of point B is:

`(x_(3), y_(3)) = \left((2)/(3)\cdot x_(2) + (1)/(3) \cdot x_(1), (2)/(3)\cdot y_(2) + (1)/(3) \cdot y_(1) \right)`