Section formula derivation plz
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mx1+nx2/m+n
my1+ny2/m+n
.
after read this u deffinately no type of prblm yo get
my1+ny2/m+n
.
after read this u deffinately no type of prblm yo get
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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.
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.
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