Here is the proof of the section formula.
Consider any two points A (x 1 , y 1 ) and B (x 2 , y 2 ) 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.
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