The Distance Formula
In the coordinate plane, Pythagorean Theorem can be used to find the distance between two points.
Let B(x_{1}, y_{1}) and A(x_{2},
y_{2}) be any two points in the coordinate plane.
From A draw a vertical line and from B draw a horizontal line.
Let the intersection of these two lines be C(x_{2}, y_{1}).
Let AB = c, CB = a = x_{2}  x_{1} and CA = b = y_{2}  y_{1}.
This result is called the distance formula. If the endpoints of a line segment in the coordinate plane
are B(x_{1}, y_{1}) and A(x_{2}, y_{2}), then:
For example: in the coordinate plane, Pythagorean Theorem can be used to find the distance between two points. Say the distance between A (1, 2) and B
(4, 6), (as
indicated in Fig: 1):
Notice thatABC is a right triangle, where
AC = 3,
And CB = 4.
Use the Pythagorean Theorem now:
AB^{2} = 3^{2} + 4^{2}
i.e. AB^{2} = 25.
Taking out the sq root,
it gives AB = 5.
It can be written in the general form, if (x_{1}, x_{2}) and (y_{1}, y_{2}) are two points in the coordinate plane; the distance between them is given by the formula as stated above:
Notice that if there are two points on a graph say like (2, 3) and (2, 6), you can find the distance between them simply by counting units on the graph, as they lie in a vertical line on the graph. Similarly you can find the distance between the points simply by counting units on the graph, if they lie on a horizontal line.
The Pythagorean Theorem, which is used to find the hypotenuse of a right triangle, can also be used in three dimensions to find the diagonal length of a rectangular prism. Say D is the distance from one corner of the box to the farthest opposite corner, with
a, b and c as length, width and height respectively,
But a more useful application of the Pythagorean Theorem in three dimensions is to find the distance between any two points in 'space'.
Look into some of the illustrative example below for sharpening the skills to help doing home work and solving real life examples related to Distance Formula.
Example 1: Given find the distance between the points (4, 6) and (2, 2).
You may use the formula
Step 1: (x_{1}, x_{2}) and (y_{1},
y_{1}) are two points i.e. (4, 6) and (2, 2) respectively.
Step 2: Distance between the two points (D) = AB = simplify it,
=
(36 + 64) = 10
Distance between the two points is 10 units, as the final answer.
Example 2: Given find the distance between the points (–3, –4) and (–5, 5).
Step 1: Plug in the values of coordinates in the distance formula.
Step 2: D =
=
{(5 + 3)^{2}+ (5 + 4)^{2}} =
(4
+ 81) =
85
= 9.2
Distance between the given points is
85
or 9.29 (rounded to first decimal), as the final answer.
Example 3: Given You have a huge rectangular box with dimensions 8 ft. x 10 ft. x 16 ft. What is the length of the longest rod you could fit in the box? The box is of given dimensions is shown in figure below:
Notice that the longest rod you could fit in the box will have distance between
A and D i.e. diagonal AD.
Step 1: To find AD, draw a rightangle triangle inside the box dimensions i.e. AD as its hypotenuse. Then apply Pythagorean Theorem.
Step 2: First find AC, which is the diagonal of rectangle.
Step 3: As ABC is a right angled triangle, so
AC^{2} = AB^{2} + BC^{2} = 16^{2} + (10)^{2} = 256 + 100 = 356.
Step 4: Then you can find AD i.e. AD^{2} = AC^{2} + CD^{2}= 356 + 8^{2} = 356 + 64 = 420
Taking sq root on both sides AD = 20.5 ft.
The longest rod you could fit in the box will be 20.5 ft, as the final answer.
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