Solutions to Distance Problems
The detailed solutions to the distance problems are presented.
Note: In what follows, sqrt means square root of.
Solution to Problem 1:
Solution to Problem 2:
Solution to Problem 3:
Solution to Problem 4:
Solution to Problem 5:
Solution to Problem 6:
Let us first define the distances D1, D2 and D3 as follows
D1 = distance from (0, 1) to (2, 3)
D2 = distance from (0, 1) to (2, -1)
D2 = distance from (2, 3) to (2, -1)
We now use the distance formula to find the above distances.
D1 = sqrt (8) , D2 = sqrt (8) and D3 = 4
Note that two sides of the triangle have equal lengths so the triangle is isosceles. For the triangle to be a right triangle, D3 the largest of the three distances must be the length of the hypotenuse and all three sides must satisfy Pythagora's theorem. Let us find the sum of the squares of D1 and D2 and the square of D3
(D1) 2 + (D2) 2 = sqrt (8) 2 + sqrt (8) 2 = 16
(D3) 2 = 4 2 = 16
We can write
(D1) 2 + (D2) 2 = (D3) 2
D1, D2 and D3 satisfy Pythagora's theorem and therefore the triangle, whose vertices are the three points given above, is a right and isosceles triangle.
Solution to Problem 7:
Solution to Problem 8:
Solution to Problem 9:
Solution to Problem 10:
More math problems with detailed solutions in this site.