Question

Find the distance between the points (x + y, x – y) and (x – y, x + y).
(a) 0
(b) $\sqrt{2}xy$
(c) $2\sqrt{2}y$
(d) $2xy$

Answer 1

Hint: First, start by defining the distance formula: $d=\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}$. Then find the relation between the distance formula and the Pythagoras theorem to understand it better. Next, use the distance formula for the given points: (x + y, x – y) and (x – y, x + y). solve and arrive at the final answer.

Complete step-by-step solution:
In this question, we need to find the distance between the points (x + y, x – y) and (x – y, x + y).
To solve this question, we will use the distance formula.
Let us first define what distance formula actually is.
The distance formula is used to determine the distance, d, between two points. If the coordinates of the two points are (a, b) and (c, d), the distance equals the square root of the sum of the squares of (a – c) and (b – d).
i.e. $d=\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}$
The Distance Formula is a variant of the Pythagorean Theorem. The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points.
So, using the distance formula for the points (x + y, x – y) and (x – y, x + y), we will get the following:
$d=\sqrt{{{\left( x+y-\left( x-y \right) \right)}^{2}}+{{\left( x-y-\left( x+y \right) \right)}^{2}}}$
$d=\sqrt{{{\left( x+y-x+y \right)}^{2}}+{{\left( x-y-x-y \right)}^{2}}}$
$d=\sqrt{{{\left( 2y \right)}^{2}}+{{\left( 2y \right)}^{2}}}$
$d=\sqrt{2\times {{\left( 2y \right)}^{2}}}$
$d=2y\times \sqrt{2}$
$d=2\sqrt{2}y$.
So, $2\sqrt{2}y$ is the distance between the points (x + y, x – y) and (x – y, x + y).
Hence, option (c) is correct.

Note: In this question, it is very important to know about the distance formula and how it is derived. If we are given two points, we draw a straight horizontal line from one point to the abscissa of the other point. Similarly, we draw a straight vertical line from the second point to the ordinate of the first point. Then join the two points together. We will get a right angled triangle and the hypotenuse is the required distance. This is how the distance formula relates to the Pythagoras theorem.