Problems
Problem 1:
Find the distance between the points (2, 3) and (0, 6).
Problem 2:
Find the distance between point (1, 3) and the midpoint of the line segment joining (2, 4) and (4, 6).
Problem 3:
Find x so that the distance between the points (2, 3) and (3, x) is equal to 5.
Problem 4:
Find x and y if (2, 5) is the midpoint of points (x, y) and (5, 6).
Problem 5:
Find the point (0, y) that is equidistant from (4, 9) and (0, 2).
Problem 6:
Show that the triangle that has (0, 1), (2, 3) and (2, 1) as vertices is right isosceles.
Problem 7:
Find the length of the hypotenuse of the right triangle whose vertices are given by the points (2, 1), (1, 1) and (1, 2).
Problem 8:
Find a relationship between x and y so that the distance between the points (x, y) and (2, 4) is equal to 5.
Problem 9:
Find a relationship between x and y so that (x, y) is equidistant from the two points (3, 4) and (0, 3).
Problem 10:
Find a relationship between x and y so that the triangle whose vertices are given by (x, y), (1, 1) and (5, 1) is a right triangle with the hypotenuse defined by the points (1, 1) and (5, 1).
Solutions to the Above Problems
Solution to Problem 1:
The formula for the distance D between two points (a, b) and (c, d) is given by
D = √ [ (c  a)^{ 2} + (d  b)^{ 2} ]
Apply the formula given above to find distance D between the points (2, 3) and (0, 6) as follows
D = √ [ (0  2)^{ 2} + (6  3)^{ 2} ]
= √ (13)
Solution to Problem 2:
We first find the coordinates of the midpoint M of the segemnt joining (2, 4) and (4, 6)
M = [ (2 + 4) / 2 , (4 + 6) / 2 ]
= (3, 5)
We now use the distance formula to find the distance between the points (1, 3) and (3, 5)
D = √ [ (3  (1))^{ 2} + (5  (3))^{ 2} ]
= √ (80) = 4 √ (5)
Solution to Problem 3:
Use the distance formula to write an equation in x.
5 = srqt [ (3  (2))^{ 2} + (x  (3))^{ 2} ]
Simplify the expression under the square root
5 = srqt [ 1 + (x + 3)^{ 2} ]
Square both sides
25 = 1 + (x + 3)^{ 2}
Solve for x
(x + 3)^{ 2} = 24
two Solutions: x = 3 + 2 √ (6) and x = 3  2 √ (6)
Solution to Problem 4:
Use midpoint formula.
(2, 5) = [ (x + (5)) / 2 , (y + 6) / 2 ]
Equate the coordinates
2 = (x  5) / 2 and 5 = (y + 6) / 2
Solve for x and y
x = 9 and y = 4
Solution to Problem 5:
Use distance formula to find distance the D1 from (0, y) to (4, 9) and the distance D2 from (0, y) to (0, 2).
D1 = √ [ (4  0)^{ 2} + (9  y)^{ 2} ]
D2 = √ [ (0  0)^{ 2} + (2  y)^{ 2} ]
The two distances are equal, hence
√ [ (4  0)^{ 2} + (9  y)^{ 2} ] = √ [ (0  0)^{ 2} + (2  y)^{ 2} ]
Square both sides and simplify
16 + (9  y)^{ 2} = (2  y)^{ 2}
16 + 81 + 18 y + y^{ 2} = 4 + y^{ 2} + 4 y
Simplify and solve for y.
y = 93 / 14
The point that equidistant from the two given points is given by
(0 , 93 / 14)
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 = √ (8) , D2 = √ (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 Pythagorean theorem. Let us find the sum of the squares of D1 and D2 and the square of D3
(D1) ^{ 2} + (D2) ^{ 2} = √ (8) ^{ 2} + √ (8) ^{ 2} = 16
(D3) ^{ 2} = 4 ^{ 2} = 16
We can write
(D1) ^{ 2} + (D2) ^{ 2} = (D3) ^{ 2}
D1, D2 and D3 satisfy Pythagorean theorem and therefore the triangle, whose vertices are the three points given above, is a right and isosceles triangle.
Solution to Problem 7:
Let D1 be the distance from (2, 1) to (1, 1), D2 be the distance from (2, 1) to (1, 2) and D3 the distance from (1, 1) to (1, 2). We now use the distance formula to find these distances.
D1 = √ [ (1  2)^{ 2} + (1  1)^{ 2} ] = 3
D2 = √ [ (1  2)^{ 2} + (2  1)^{ 2} ] = √ (10)
D3 = √ [ (1  1)^{ 2} + (2  1)^{ 2} ] = 1
If the triangle defined by the three vertices is a right triangle, the hypotenuse is the side with the largest size and that is D2 = √ (10). We now use Pythagorean theorem to check that it is a right triangle with D2 the hypotenuse and D1 and D3 the sides of the triangle.
(D1) ^{ 2} + (D3) ^{ 2} = 3^{ 2} + 1^{ 2} = 10
(D2) ^{ 2} = √(10) ^{ 2} = 10
According to the above, we can write
(D1) ^{ 2} + (D3) ^{ 2} = (D2) ^{ 2}
The triangle defined by the given points is a right triangle and the size of it hypotenuse is √(10).
Solution to Problem 8:
We apply the distance formula.
5 = √ [ (2  x)^{ 2} + (4  y)^{ 2} ]
Square both sides.
25 = (2  x)^{ 2} + (4  y)^{ 2}
The above relationship between x and y is the equation of a circle. As an exercise explain why.
Solution to Problem 9:
Let D1 be the distance from (x, y) to (3, 4) and find it.
D1 = √ [ (3  x)^{ 2} + (4  y)^{ 2} ]
Let D2 be the distance from (x, y) to (0, 3) and find it.
D2 = √ [ (0  x)^{ 2} + (3  y)^{ 2} ]
The two distances are equal, hence.
√ [ (3  x)^{ 2} + (4  y)^{ 2} ] = √ [ (0  x)^{ 2} + (3  y)^{ 2} ]
We now square both sides and simplify.
(3  x)^{ 2} + (4  y)^{ 2} = (0  x)^{ 2} + (3  y)^{ 2}
9 + x^{ 2} + 6 x + 16  8 y + y^{ 2} = x^{ 2} + 9 + y^{ 2} + 6y
6x  14 y + 16 = 0
The above relationship between x and y is the equation of a line which is the perpendicular bisector of the line segment defined by the points (3, 4) and (0, 3).
Solution to Problem 10:
Let us use the distance formula to find the length of the hypotenuse h.
h = √ [ (5  1)^{ 2} + (1  1)^{ 2} ] = 4
We now use the distance formula to find the sizes of the two other sides a and b of the triangle.
a = √ [ (x  1)^{ 2} + (y  1)^{ 2} ]
b = √ [ (x  5)^{ 2} + (y  1)^{ 2} ]
Pythagorean theorem gives
4^{ 2} = (x  1)^{ 2} + (y  1)^{ 2} + (x  5)^{ 2} + (y  1)^{ 2}
Expand the squares, simplify and complete the squares to rewrite the above relationship between x and y as follows.
(x  3)^{ 2} + (y  1)^{ 2} = 2^{ 2}
The above relationship between x and y is the equation of a circle. Explain why.
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