Question

a particle starting from point A(1,2) moves 2√2 m along the straight line x-y+1=0.Then final position vector of the particle

Answer 1

Answer:

There can be two possible points:

(-1,0) & (3,4)

Explanation:

Let us suppose the point be (a,b);

∴From the distance formula,

  $(a-1)^{2} + (b-2)^{2} = 8$

  $a^{2} -2a+1 + b^{2} - 4b +4 = 8\\a^{2} + b^{2} -2a -4b = 3\\Also,\\a-b+1 = 0\\a-b = -1\\a = b-1\\So,\\(b-1)^{2} + b^2 -2(b-1)-4b = 3\\b^2 - 2b + 1+b^2 -2b +1 -4b = 3\\2b^2 -8b =0\\b^2 - 4b = 0\\b(b-4)=0\\So,\\b = 0, 4\\a = -1, 3\\So, the points are (-1,0) & (3,4)$