Write proof for the section formula.
A section formula can be either applied internally or externally.
Answer: Section formula for division of coordinates (\(x_{2}\), \(y_{2}\)) and ( \(x_{1}\), \(y_{1}\) ) in the ratio m : n, will be M(x, y) = ( \(mx_{2}\) + \(nx_{1}\) )/(m + n), ( \(my_{2}\) + \(ny_{1}\) )/(m + n).
Let us go through the step-by-step explanation to understand the proof in detail.
Explanation:
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)