- #1
solakis1
- 422
- 0
prove:
\(\displaystyle \sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}\)
\(\displaystyle \sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}\)
solakis said:prove:
\(\displaystyle \sqrt{a^2+b^2}\leq\sqrt{(x-a)^2+(y-b)^2}+\sqrt{x^2+y^2}\)
kaliprasad said:This comes from law of triangle sides
distance from origin to (a,b) < distance from (x,y) to (a,b) + distance from origin to (x,y) and equal if (x,y) is between (0,0) and (a,b) and in the line between the 2
solakis said:Can you draw a picture ??
I was meaning an algebraic solution in my OP
The equation is a representation of the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Yes, the proof involves using the properties of inequalities and the Pythagorean Theorem. We can also visualize the concept by drawing a right triangle and manipulating the sides to show that the equation is true.
The equation is useful in various fields of science, such as physics and engineering, where calculations involving distances and lengths are common. It allows us to determine the maximum distance between two points on a Cartesian plane and is also applicable in solving optimization problems.
No, the equation holds true for all values of a, b, x, and y. This is because the Pythagorean Theorem is a fundamental principle in mathematics and cannot be violated.
The equation has many practical applications, such as determining the shortest distance between two points on a map or calculating the length of a diagonal in a rectangular room. It is also used in various fields of engineering, such as in construction and navigation.