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.

Best Wishes @!

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