A) \[\frac{nK-m}{K-1}\]
B) \[\frac{K-m}{K-n}\]
C) \[\frac{K+m}{K+n}\]
D) \[\frac{1+K}{m+nK}\]
Correct Answer: A
Solution :
(a) \[K=\frac{x-m}{x-n}\] \[\Rightarrow \]\[K(x-n)=x-m\] \[\Rightarrow \]\[Kx-nK=x-m\] \[\Rightarrow \,\,x=\frac{nK-m}{K-1}\] \[\Rightarrow (k-1)x\] \[=nK-m\]
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