What is the Distance Formula?
The distance formula is used to find the distance between any two points on a plane when we know the coordinates of these two points. These coordinates may be located on the x or y axis, or even on both. The x-coordinate, also known as the abscissa, refers to a distance of a point from the y-axis. The y-coordinate, also called the ordinate, is the distance of a point from the x-axis. Consider an XY plane with two points A and B. \( (x_1,~y_1) \) are the coordinates of point A, and \( (x_2,~y_2) \) are the coordinates of point B.
So the formula to determine the distance between the points AB is:
\( AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
In this image ‘d’ is the distance between A and B.
[Note: The distance between two points is always positive.]
Rapid Recall
Solved Examples
Example 1: In a 2D plane, there are two points P and Q with the coordinates (3, 4) and (-5, 7) respectively. What will be the distance between PQ?
Solution:
As stated in the question, two points P and Q have the coordinates (3, 4) and (-5, 7).
Then,
\( P(3,~4) = P (x_1,~y_1) \)
\( Q(-5,~7) = P (x_2,~y_2) \)
Now, the distance formula between P and Q is written as,
\( PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( ~~~~~~=\sqrt{(-5-3)^2+(7-4)^2} \)
\( ~~~~~~=\sqrt{(-8)^2+(3)^2} \)
\( ~~~~~~=\sqrt{64+9} \)
\( ~~~~~~=\sqrt{73} \)
\( ~~~~~~\approx 8.5 \)
Therefore, the distance between points P and Q is 8.5 units.
Example 2: Adam’s school is located 5 miles north, 3 miles east of his apartment. Adam picks up his friend John, who lives 4 miles west and 7 miles north of Adam’s house, every day as he travels to school. Calculate the distance of John’s home from their school.
Solution:
Let’s assume the coordinates for Adam’s apartment are (0, 0).
Then, let A be the location of Adam’s school with coordinates A (3, 5).
And, John’s home is located at point B with the coordinates B (-4, 7).
By using the distance formula, we can find the distance between A and B.
That is,
\( AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( ~~~~~~=\sqrt{(-4-3)^2+(7-5)^2} \)
\( ~~~~~~=\sqrt{(-7)^2+(2)^2} \)
\( ~~~~~~=\sqrt{49+4} \)
\( ~~~~~~=\sqrt{53} \)
\( ~~~~~~\approx 7.28 \)
Hence, the distance between John’s house and their school is 7.28 miles.
Example 3: Anna wants to buy a dress in the nearby boutique. Anna’s house is located at the center of the city and the boutique is situated 2 miles south and 6 miles east from Anna’s house. Find the distance between Anna’s house and the boutique.
Solution:
It is stated that Anna’s house is at the center of the city. That is, the coordinates for it will be A (0, 0). As stated, the coordinates of the boutique are B (6, -2).
Then, we can determine the distance from Anna’s house to the boutique with the help of the distance formula.
\( AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( ~~~~~~=\sqrt{(6-0)^2+(-2-0)^2} \)
\( ~~~~~~=\sqrt{(6)^2+(-2)^2} \)
\( ~~~~~~=\sqrt{36+4} \)
\( ~~~~~~=\sqrt{40} \)
\( ~~~~~~\approx 6.32 \)
Thus, the distance between Anna’s house and the boutique is 6.32 miles.
In physics, distance is related to the speed and hence, the distance traveled by a person or a vehicle. But in math, distance refers to the separation between two coordinate points on a plane.
Just like the Pythagorean theorem, here the two points \( (x_1,~y_1) \) and \( (x_2,~y_2) \) acts as the two vertices of a right-angled triangle that when connected make up the hypotenuse.
We can find the distance between two points by using the distance formula.
Although the coordinates may have negative values, the distance between two points should always be positive.