Find the value of K when the distence between the point (3,2k) and (4,

1.	Find the value of K when the distence between the point (3,2k) and (4,1) is √10 unit .
Answer» Using distant formula√^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>=√10Square both side then1+1+4k^²-4k=104k^²-4k=84(k^2-k)=8K^²-k-2=0K^²-2k+k-2=0K(k-2) 1(k-2) =0(K+1) (k-2)=0(K+1)=0 or (k-2)=0K=-1 or k=+2

Discussion