Question

derive section formula and give the exact answer

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 y2 - 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 + nThe 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 1m+n

Answer 2

Section formula:-
 

The coordinates of the point which divides the line segment joining

(x1,y1) and (X2,y2)

internally in the ratio of m:n is

(mx2+nx1/m+n, my2+ny1/m+n)