Mid-point ratio on graph

Question

I have two points $(4,-5)$ and $(0,7)$ and I'm supposed to use the ratio $1:3$ to place it on the grid. Could you please show both the normal equation and one using the example coordinates?

Answer 1

If $A = (4,-5)$, and $B = (0,7) $, then a point $P = (x, y) $ on $AB$ is given by

$ P = (1 - t) A + t B $ where $ 0 \le t \le 1 $

Note that

$AP = P - A = - t A + t B = t (B - A) $ and

$PB = B - P = (1 - t)A + (t - 1) B = (t - 1) (B - A) $

So we want

$ \dfrac{t}{|t - 1|} = \dfrac{t}{1 -t } = \dfrac{1}{3} $

From here, it follow that $ 3 t = 1 - t$ which give us $ t = \dfrac{1}{4} $

Substituting this, we get

$ P = \left(\dfrac{3}{4}\right) (4, -5) + \left(\dfrac{1}{4}\right) (0, 7) = ( 3 , \dfrac{ -15 + 7 }{4} ) = \boxed{ (3, -2)} $