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Statement -1 : If a transversal cuts t...

Statement -1 : If a transversal cuts the sides OL, OM and diagonal ON of a parallelogram at A, B, C respectively, then
`(OL)/(OA) + (OM)/(OB) =(ON)/(OC)`
Statement -2 : Three points with position vectors ` veca , vec b , vec c ` are collinear iff there exist scalars x, y, z not all zero such that `x vec a + y vec b +z vec c = vec 0, " where " x +y + z=0.`

Text Solution

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We have,
ON=OL+LN=OL+OM` . . (i) ltb rgt Let `OL=xOA,OM=yOB` . . . (ii)
annd ON=zOC
so, `|OL|=x|OA|,|OM|=y|OB| and |ON|=z|OC|`
`thereforex(OL)/(OA),y=(OM)/(OB) and z=(ON)/(OC)`
`therefore` From Eqs. (i) and (ii), we have

`zOC=xOA+yOB`
`implies xOA+yOB-zOC=0`
`therefore`Points A,B and C are collinear, the sum of the coefficients of their PV must be zero.
`implies x+y-z=0`
i.e., `(OL)/(OA)+(OM)/(OB)=(ON)/(OC)`
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