he line segment joining thepoints ( 3, 4) and ( K, 2 ) is divided by the line 2x+3y+5=0in the ratio 2:3 then K=

CLEARLY BY APPLYING SECTION FORMULA WE GET (X,Y)=[{(2K+9)/5},{(4+12)/5}]

SO FROM COMPARING X={(2K+9)/5}, AND Y=16/5

 SINCE C(X,Y) LIES ON THE LINE 2X+3Y+5=0 IT SHOULD SATISFY THE EQUATION SO SUBSTITUTE THE VALUES OF ''X'' AND ''Y'' AND GET THE VALUE OF ''K''

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Dear Akhil,

The equation of the straight line joining the points (6, 8) and (-3, -2) is
y -8 = [(-2 -8)/(-3 -6)](x -6)                                  (Two-point form)
=> y -8 = (10/9) (x -6)   =>  9 y -72 = 10 x -60
=> 10 x -9 y +12 = 0         ...(i)
Let the line joining the points (6, 8) and (-3, -2) i.e. the line (i) divide the line segment joining the points (2, 3) and (4, -5) at the point P in the ratio k : 1, then the co-ordinates of point P are
((k.4 +1.2)/(k+1), (k (-5) +1 .3)/(k+1)) i.e. ((4k +20)/(k+1), (-5k +3)/(k+1))
Since P lies on (i), we get
10[(4k +20)/(k+1)] -9[(-5k +3)/(k+1)] + 12 = 0
=> 40 k +20 +45 k -27 +12 k +12 = 0
=> 97 k +5 = 0 => k = -5/97
Hence the required ratio is -5/97 i.e. 5 : 97 externally.

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