1please solve the problem,nishant sir.
Find th eq. of the locus of the point which moves such that its distance frm x-y+1+0 is twice its distance from x+y+6=0.
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3 Answers
62Take the point as h,k
the distance from 1st line is twice from second
straight formula of distance of a point from a line..
THat is what you have to do.
1pura bata to diya hai phir bhi since u r asking me posting the process
let point be (h,k)
according to given condition
2\left|\frac{h-k+1}{\sqrt{h^{2}+k^{2}}} \right|=\left|\frac{h+k+6}{\sqrt{h^{2}+k^{2}}} \right|
here LHS is distance of point (h,k) from x-y+1=0 ( note that 2 in LHS is in accordance with condition the point which moves such that its distance frm x-y+1+0 is twice its distance from x+y+6=0. )
and RHS is distance of (h,k) from x+y+6=0
now solving abv u get
h - 3k - 4 = 0
hence locus is x - 3y - 4 =0