Question

22. Using section formula ,prove that the three points (-2,3,5),(1,2,3) and (7,0,-1) are
collinear. Also find the ratio in which the third point divides the segment joining the first
two points.
23. Find the equation of the locus of all points such that difference of their distance from
(4,0)&(-4,0 is always 2.​

Answer 1

Answer:

22. Given points are A(-2,3,5), B(1,2,3) and C(7,0,-1)

Points A,B,C, are collinear if point" c" divides the AB in ratio externally or internally.

By section formula we have

(x,y,z) = (mx2+nx1/m+n, my2+my1/m+n mz2+nz1/m+n)

here, Let the point B(1,2,3) divide the AC in the ratio of k:1

now,

m= k and n = 1 and x1 =-2, y1 = 3, z1 = 5, x2 =1, y2 = 2 and z2 = 3

therefore

by using d=section formula we get B( 7k-2/k+1, 3/k+1, -k+5/k+1) --(1)

now comparing the (1) from co-ordinate of B.

we get  7k-2/k+1 = 1, therefore k = 1/2

3/k+1 = 2 therefore k 1/2 and -k+5/k+1 =3 , therefore k =  1/2

hence we get the ratio 1:2 and it divide the line segment AB in ratio of 1:2.

Hence the given points are collinear and the ratio will be 1:2

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