The distance formula is a formula that is used to find the distance between two points. These points can be in any dimension. For example, you might want to find the distance between two points on a line (1d), two points in a plane (2d), or two points in space (3d).
Suppose and are two points lying on the real number line. Then the distance between and is
In the plane, we can consider the -axis as a one-dimensional number line, so we can compute the distance between any two points lying on the -axis as the absolute value of the difference of their -coordinates. Similarly, the distance between any two points lying on the -axis is the absolute value of the difference of their -coordinates.
Now, consider the -plane, and suppose and are two points in it . Then the distance between and is
Since is the distance between the -coordinates of the two points and is the distance between the -coordinates of the two points, the distance formula in the -plane can be thought of as the length of the hypotenuse of the right triangle with vertices , and . Then the distance formula is simply a statement of the Pythagorean theorem.
In both 1D and 2D, the distance function satisfies the following properties:
- for all points with equality if and only if
- for all points
- for all points .
What is the distance between the points and ?
Note that both of these points lie on the -axis and therefore the distance between the points is the absolute value of the difference of the -coordinates, which is
To generalize the above problem, if two points and have the same -coordinate, i.e. , then the distance between the two points is and the line segment is a vertical line segment.
Similarly, if and have the same -coordinate (), then and the line segment is a horizontal line segment.
Find the area of the rectangle in the -plane with vertices
Points and have the same -coordinate, implying . Points and have the same -coordinate, implying . We check that points and have the same -coordinate and and have the same -coordinate, implying the points are indeed vertices of a rectangle.
The area of the rectangle is then
The distance between two points and can be found using the following formula:
Construct a triangle where has the coordinates .
Then is a right angled triangle, and we can apply the Pythagorean theorem to obtain
Since is to be found, and and we have
From this proof we can derive tbe following corollary:
The distance of point from the origin is given by
What is the distance between the two points and
The distance is
Find the sum of all such that the distance between the points and is .
Using the above formula we get
Squaring both sides and on simplifying, we get
Therefore the sum of possible values of is
Find the distance between the points and .
Find the distance of the point from the origin.
Sometimes we are given four points and asked to comment on the nature of the quadrilateral which is formed by joining them. For this, we have to recall the following:
A quadrilateral is a
- rectangle, if its opposite sides are equal and diagonals are equal;
- square, if all its sides are equal and diagonals are equal;
- parallelogram, if its opposite sides are equal;
- rhombus, if its sides are equal.
Show that the points are the vertices of an isosceles right-angled triangle. Also find the area of the triangle.
Since the triangle is isosceles.
Moreover, since it is right-angled.
Now, the area of the triangle is
Show that the points are the vertices of a rectangle. Also find the area of the rectangle.
We have which implies and i.e. is a quadrilateral whose opposite sides are equal.
Now, since we have which implies is a quadrilateral whose diagonals are equal.
Hence, is a rectangle, and its area is
Show that four points in a plane form a rhombus which is not a square. Find the area of the rhombus.
We have which implies that i.e. is either a rhombus or a square.
Now, since we have which implies that is a rhombus but not a square.
The area of rhombus is
Show by the distance formula that are collinear.
We have so which implies that the given points are collinear.
Find a point on the -axis which is equidistant from the points and .
Since the point lies on the -axis, the -coordinate is . Let be the required point on the -axis which is equidistant from the given points. Then,
Hence, the required point is