# 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 $P \left(x , y\right)$ lies on the interval $A \left({x}_{1} , {y}_{1}\right) , B \left({x}_{2} , {y}_{2}\right)$ such that $A P : P B = a : b$, with a and b positive,show that $x = \frac{b {x}_{1} + a {x}_{2}}{b + a}$ and $y = \frac{b {y}_{1} + a {y}_{2}}{b + a}$? .

Aug 27, 2016

See below.

#### Explanation:

If $P \in \overline{A B}$ then

$P = A + \lambda \left(B - A\right) , \lambda \in \left[0 , 1\right]$ or equivalently

$P = B + \mu \left(A - B\right) , \mu \in \left[0 , 1\right]$

also

$\left\lVert P - A \right\rVert = \lambda \left\lVert B - A \right\rVert$ and
$\left\lVert P - B \right\rVert = \mu \left\lVert A - B \right\rVert$

but

$\frac{\left\lVert P - A \right\rVert}{\left\lVert P - B \right\rVert} = \frac{\lambda}{\mu} = \frac{a}{b} = \frac{a \left(a + b\right)}{b \left(a + b\right)}$

but anyway

$\lambda = \frac{a}{a + b}$ and $1 - \lambda = \frac{b}{a + b}$

$P = \left(1 - \lambda\right) A + \lambda B = \frac{b}{a + b} A + \frac{a}{a + b} B$

resulting in

$x = \frac{b {x}_{1} + a {x}_{2}}{a + b}$

$y = \frac{b {y}_{1} + a {y}_{2}}{a + b}$

Aug 27, 2016

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

#### Explanation:

From the figure solve for x:

$a {x}_{2} - a {x}_{1} = \left(a + b\right) \left(x - {x}_{1}\right) = a x - a {x}_{1} + b x - {x}_{1} = \left(a + b\right) x - \left(a + b\right) {x}_{1}$
Rearranging and solving for x we have:
$x = \frac{a {x}_{2} + b {x}_{1}}{a + b}$

Applying the same logic to the ratio:
$\frac{a}{a + b} = \frac{\overline{P C}}{\overline{B D}} = \frac{y - {y}_{1}}{{y}_{2} - {y}_{1}}$ Solving for $y$:
$y = \frac{a {y}_{2} + b {y}_{1}}{a + b}$