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 (x2, y2) and (x1, y1) in the ration m : n, will be M(x, y) = (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)
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