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`        proof of section formula in straight line , where a point devides line sigment in m:n ratio `
2 years ago

```							Dear Ankit  Consider ? ABB’  Since BB’||PQ and AP:PB = m:n (see figure given below) AQ/AB' = PQ/BB' = m/(m+n) (= AP/AB) ? x – x1/x2 – x1 = m/(m+n) ? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) If P is outside AB (less assume it is at P’) We have (x – x1)/(x2 – x1) = m/(m + n) ? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) Similarly if P is at P” then x = (–mx2 + nx1)/(n-m), y = (– my2 + ny1)/(n-m)Note:  m:n can be written as m/n or ?:1. So any point on line joining A and B will be P{(?x2 + x1)/(? + 1), (?y2 + y1)/(? +1)}. It is useful to assume ?:1 because it involves only one variable.  RegardsArun (askIITians forum expert)
```
2 years ago
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