Question

If `P(x,y)` lies on the interval `A(x_1,y_1), B(x_2,y_2)` such that `AP : PB =a : b`, with a and b positive, show that `x= (bx_1+ax_2) /(b+a)` and `y=(by_1+ay_2)/(b+a)`?

If `P(x,y)` lies on the interval `A(x_1,y_1), B(x_2,y_2)` such that `AP : PB =a : b`, with a and b positive,show that
`x= (bx_1+ax_2) /(b+a)`
and `y=(by_1+ay_2)/(b+a)`?
.

Answer 1

See below.

Explanation:

If `P in bar(AB)` then

`P =A +lambda (B-A), lambda in [0,1]` or equivalently

`P = B + mu(A-B), mu in [0,1]`

also

`norm(P-A) = lambda norm(B-A)` and
`norm(P-B) = mu norm(A-B)`

but

`norm(P-A)/norm(P-B) = lambda/mu = a/b = (a(a+b))/(b(a+b))`

but anyway

`lambda = a/(a + b)` and `1-lambda = b/(a+b)`

so another reading is

`P = (1-lambda)A+lambda B=b/(a+b)A+a/(a+b)B`

resulting in

`x = (b x_1+a x_2)/(a+b)`

`y = (b y_1+a y_2)/(a+b)`

Answer 2

See the geometric construction to guide you with set up...

Explanation:

From the figure solve for x:

`ax_2-ax_1=(a+b)(x-x_1)=ax-ax_1+bx-x_1=(a+b)x-(a+b)x_1`
Rearranging and solving for x we have:
`x= (ax_2+bx_1)/(a+b)`

Applying the same logic to the ratio:
`a/(a+b) = bar(PC)/bar(BD)= (y-y_1)/(y_2-y_1)` Solving for `y`:
`y = (ay_2+by_1)/(a+b)`