would any1 mind giving me the proof of the section formula????????
Hi!
Here is the proof of the section formula.
Consider any two points A (x1, y1) and B (x2, y2) and assume that P (x, y) divides AB internally in the ratio m: n i.e. PA: PB = m: n
Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC perpendiculars to PS and BT respectively.
In ∆PAQ and ∆BPC
∠PAQ = ∠BPC (pair of corresponding angles)
∠PQA = ∠BCP (90°)
Hence, ∆PAQ ∼ ∆BPC (AA similarity criterion)
Hope! You got the concept.
Cheers!