Question

derivation of section formula in co-ordinate geometry

Answer 1

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) ∴ PA BP = AQ PC = PQ BC (Corresponding sides of similar triangle are proportional) ⇒ m n = x - x 1 x 2 - x = y - y 1 y 2 - y Taking m n = x - x 1 x 2 - x , we get m x 2 - mx = nx - n x 1 ⇒ (m+n)x = m x 2 + n x 1 ⇒ x = m x 2 + n x 1 m + n Similarly taking m n = y - y 1 y 2 - y , we get y = my 2 + n y 1 m + n The coordinates of the points P(x, y) which divides the line segment joining the points A( x 1 , y 1 ) and B( x 2 , y 2 ), internally, in ratio m : n are m x 2 + n x 1 m+n , m y 2 + n y 1 m+n