# Section Formula

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

Let A(x1, y1) and B(x2, y2) 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 - x1

PT = MN = ON - OM = x2 - x

KP = MP - MK = MP - LA = y - y1

TB = NB - NT = NB - MP = y2 - 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

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