A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: C
Solution :
Solving \[y={{m}_{r}}x\]and \[x+y=1\], we get \[x=\frac{1}{1+{{m}_{r}}}\] and \[y=\frac{{{m}_{r}}}{1+{{m}_{r}}}\]. Thus the points of intersection of the three lines on the transversal are \[\left( \frac{1}{1+{{m}_{1}}},\frac{{{m}_{1}}}{1+{{m}_{1}}} \right),\] \[\left( \frac{1}{1+{{m}_{2}}},\frac{{{m}_{2}}}{1+{{m}_{2}}} \right)\] and\[\left( \frac{1}{1+{{m}_{3}}},\frac{{{m}_{3}}}{1+{{m}_{3}}} \right)\] By hypothesis, \[{{\left( \frac{1}{1+{{m}_{1}}}-\frac{1}{1+{{m}_{2}}} \right)}^{2}}+{{\left( \frac{{{m}_{1}}}{1+{{m}_{1}}}-\frac{{{m}_{2}}}{1+{{m}_{2}}} \right)}^{2}}\] = \[{{\left( \frac{1}{1+{{m}_{2}}}-\frac{1}{1+{{m}_{3}}} \right)}^{2}}+{{\left( \frac{{{m}_{2}}}{1+{{m}_{2}}}-\frac{{{m}_{3}}}{1+{{m}_{2}}} \right)}^{2}}\] Þ \[\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}}=\frac{{{m}_{3}}-{{m}_{2}}}{1+{{m}_{3}}}\]or \[\frac{1+{{m}_{2}}}{1+{{m}_{1}}}-1=1-\frac{1+{{m}_{2}}}{1+{{m}_{3}}}\] Þ \[\frac{1+{{m}_{2}}}{1+{{m}_{1}}}+\frac{1+{{m}_{2}}}{1+{{m}_{3}}}=2\]Þ \[1+{{m}_{2}}=\frac{2(1+{{m}_{1}})(1+{{m}_{3}})}{(1+{{m}_{1}})+(1+{{m}_{3}})}\] Þ \[1+{{m}_{1}},1+{{m}_{2}},1+{{m}_{3}}\] are in H.P.
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