Question

The distance between points $$\left( {a + b,b + c} \right)$$ and $$\left( {a - b,c - b} \right)\;$$ is $$2\sqrt 2 b$$

A) $$2\sqrt {{a^2} + {b^2}} $$
B) $$2\sqrt {{b^2} + {c^2}} $$
C) $$2\sqrt 2 b$$
D) $$\sqrt {{a^2} - {c^2}} $$

Answer 1

Hint:
we have the formula to calculate distance between two points which is Distance between two points $$({x_1},{y_1}) and ({x_{2,}}{y_2})$$ can be calculated using the formula $$\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $$. We already have values of $$({x_1},{y_1})and({x_{2,}}{y_2})$$. After putting values of $$({x_1},{y_1})and({x_{2,}}{y_2})$$ we can get the answer.

Complete step by step solution:
Using formula to calculate distance between two points which is Distance between two points $$({x_1},{y_1})and({x_{2,}}{y_2})$$
 Fórmula= $$\sqrt{{{({x_2} - {x_1})}^{2}}+{{({y_2} - {y_1})}^{2}}}$$.
Distance between the points
 $$\eqalign{
  & \left( {a + b,b + c} \right){\text{ }}and{\text{ }}\left( {a - b,c - b} \right) = \sqrt {{{(a - b - a - b)}^2} + {{(c - b - b - c)}^2}} \cr
  & = \sqrt {4{b^2} + 4{b^2}} \cr
  & = 2\sqrt 2 b \cr} $$

Hence, correct answer is option C. $$2\sqrt 2 b$$

Note:
There can be many applications of distance formula. We can show tough concepts like collinearity easily with the help of distance formula. They can give 3 points and ask whether a triangle is possible with the given vertices. In that case we know we’ll use “Sum of two sides of a triangle is always greater than the third one” but to get the distance we can use distance formula.