Question

Derive Section Formula of Co-ordinate Geometry in an easy way ?

Answer 1

DERIVATION

Draw the figure in the first quadrant:

AB a line is divided by point P

AP = m

PB = n

Draw Δ APO taking O as any point such that AO is parallel to x-axis.

Draw Δ BPQ taking Q as any point such that BQ is parallel to y-axis.

Let the two Δs be Δ APO and Δ BPQ

[ O and Q are the base points  ]

Now this figure is very bad !

I suppose you draw this figure in copy with additional details :

Extend AO and PQ to meet at a point E .

In the two Δ s ABE and Δ APO :

∠ AOP = ∠ BEA [ 90° each ]

∠ PAO = ∠ BAE [ common ]

Hence Δ PAO ≈ Δ BAE [ A.A criteria ]

AE / AO = AB / AP [ corresponding side ratio of two similar triangles ]

See the figure very carefully :

AE = x₂ - x₁

AO = x - x₁

AP = m

AB = m + n

AE / AO = AB / AP

= > ( x₂ - x₁ ) / ( x - x₁ ) = ( m + n ) / m

Cross multiply :

= > m ( x₂ - x₁ ) = ( m + n )( x - x₁ )

= > m ( x₂ - x₁ ) / ( m + n ) = x - x₁

= > x = m ( x₂ - x₁ ) / ( m + n ) + x₁

= > x = [ m ( x₂ - x₁ ) + x₁ ( m + n ) ] / ( m + n )

= > x = [ m x₂ - m x₁ + m x₁ + n x₁ ] / ( m + n )

= > x = [ m x₂ + n x₁ ] / ( m + n )

[ P.R.O.V.E.D ]

I am not doing for " y " just giving you a hint :

Take BE / PO = AB / AP

Take BE as y₂ - y₁

PO = y - y₁

AB = m + n

AP = m

Put the values and solve for y similarly as x.

The sectional formula will be proved !

Answer 2

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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 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