To find the co-ordinates of a point, which divides the line segment joining two points, in a given ratio internally.

Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be the two given points and P(x, y) be a point on AB which divides it in the given ratio m : n. You have to find the co-ordinates of P.

Draw the perpendiculars AL, PM, BN on OX, and, AK, PT on PM and BN respectively.

AK = LM = OM – OL = x - x_{1}

PT = MN = ON - OM = x_{2} - x

KP = MP - MK = MP - LA = y - y_{1}

TB = NB - NT = NB - MP = y_{2} - y

From similar triangles APK and PBT,

$$ \frac{AP}{PB} = \frac{AK}{PT} = \frac{KP}{TB} $$

$$ \frac{m}{n} = \frac{x - x_1}{x_2 - x} = \frac{y - y_1}{y_2 - y} $$

From the first two relations we get,

$$ mx_2 - mx = nx - nx_1 $$

$$ x(m + n) = mx_2 + nx_1 $$

$$ x = \frac{mx_2 + nx_1}{m + n} $$

Similarly, from the first and third relation we get

$$ y = \frac{my_2 + ny_1}{m + n} $$