# Write proof for the section formula.

**Solution:**

A section formula can be either applied internally or externally.

**Derivation of the Formula:**

Let A ( \(x_{1}\), \(y_{1}\)) and B ( \(x_{2}\), \(y_{2}\)) be the endpoints of the given line segment AB and C(x, y) be the point that divides AB (internally) in the ratio m:n.

Then, AC / CB = m / n

We want to find the coordinates (x, y) of C.

Now draw perpendiculars from A, C, and B parallel to Y coordinate joining at P, Q, and R on X-axis respectively.

By seeing the above diagram,

AM = PQ = OQ – OP = (x – \(x_{1}\))

CN = QR = OR – OQ = (\(x_{2}\) – x)

CM = CQ – MQ = (y – \(y_{1}\))

BN = BR – NR = (\(y_{2}\) _{ }– y)

Clearly, we can see that ∆AMC and ∆CNB are similar and, therefore, their sides are proportional by the AA similarity theorem.

AC / CB = AM / CN = CM / BN

Now substituting the values in the above relation, we get,

m / n = (x – \(x_{1}\)) / (\(x_{2}\) -x) = (y – \(y_{1}\)) / (\(y_{2}\) – y)

m / n = (x – \(x_{1}\)) / (\(x_{2}\) -x) and m / n = (y – \(y_{1}\)) / (\(y_{2}\) – y)

Solving the 1st condition,

m(\(x_{2}\) – x) = n(x – \(x_{1}\))

(m + n)x = (m\(x_{2}\) + n\(x_{1}\))

x = (m\(x_{2}\) + n\(x_{1}\)) / (m + n)

Solving the 2nd condition,

m(\(y_{2}\) – y) = n(y – \(y_{1}\))

(m + n)y = (m\(y_{2}\) + n\(y_{1}\))

y = (m\(y_{2}\) + n\(y_{1}\)) / (m + n)

Thus, we have seen the proof of the section formula.

## Write proof for the section formula.

**Summary:**

Section formula for division of coordinates (x_{2}, y_{2}) and (x_{1}, y_{1}) in the ration m : n, will be M(x, y) = (mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)

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