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Find the value of K when …

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Find the value of K when the distence between the point (3,2k) and (4,1) is √10 unit .

Posted by Chitresh Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Shabnam Star 6 years, 10 months ago

Using distant formula √<x2-x1>^² + < y2-y2>^² {x1=3,x2=4,y1=2k,y2=1 √<4-3>^² +<1-2k>^² =√10 √<1>^² + <1+4k^²-4k>=√10 Square both side then 1+1+4k^²-4k=10 4k^²-4k=8 4(k^2-k)=8 K^²-k-2=0 K^²-2k+k-2=0 K(k-2) 1(k-2) =0 (K+1) (k-2)=0 (K+1)=0 or (k-2)=0 K=-1 or k=+2

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