Coordinate geometry is one of the most exciting ideas of mathematics that provides a connection between algebra and geometry through graphs of lines and curves. This enables geometric problems to be solved algebraically and provides geometric insights into algebra. You must have studied the basics of coordinate geometry in your earlier classes and these have been summarized as follows. And after which we move on to explain the main concepts of this article, which are Distance Formula and Section Formula.

**Basics of Coordinate Geometry**

The number plane (Cartesian plane) is divided into four quadrants by two perpendicular axes called the x-axis (horizontal line) and the y-axis (vertical line). These axes intersect at a point O called the origin. The position of any point in the plane can be represented by an ordered pair of numbers (x, y). Wherein:

x – is the x-coordinate or abscissa which is the distance of a point from the y-axis

y – is the y-coordinate or ordinate which is the distance of a point from the x-axis

Let us now study about two important applications of coordinate geometry which are Distance Formula and Section Formula

**Distance Formula**

Let us now find the distance between any two points A (x_{1}, y_{1}) and B (x_{2}, y_{2}).

Draw AC and BD perpendicular to the x-axis.

A perpendicular from the point A on BD is drawn to meet it at the point E.

Then,

OC = x_{1},

OD = x_{2}

So, CD = x_{2} – x_{1} = AE

Also

DB = y_{2},

DE = AC = y_{1}

So, BE = y_{2} – y_{1}

Now, applying the Pythagoras theorem in ΔAEB, we get

AB^{2} = AE^{2} + BE^{2}

= (x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Therefore, AB =

**Note that since distance is always non-negative, we take only the positive square root.**

So, the distance between the points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) is

which is called the **Distance Formula**.

**Distance Formula Examples**

Let us try to understand it better through an example:

**Example: **Find the distance between the points A (2, -5) and B (5, -1).

**Solution:**

The co-ordinates of point A and B are (2, -5) and (5, -1).

Here,

x_{1} = 2, x_{2} = 5

And y_{1} = -5, y_{2} = -1

So, the distance between points A (2, -5) and B (5, -1) is given by

Therefore, the distance between points A (2, -5) and B (5, -1) is 5 units.

**Section Formula**

Let us try to derive the section formula now:

Consider any two points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) and assume that P (x, y) divides AB internally in the ratio m_{1}: m_{2}, i.e.,

Draw AR, PS and BT perpendicular to the *x*-axis.

Draw AQ and PC parallel to the *x*-axis.

Then, ΔPAQ ~ ΔBPC (AA similarity criterion)

Therefore,

Now,

AQ = RS = OS – OR = **x – x _{1}**

PC = ST = OT – OS = **x _{2} – x**

PQ = PS – QS = PS – AR = **y – y _{1}**

BC = BT– CT = BT – PS = **y _{2} – y**

Substituting these values in (i), we get

Taking,

So, the coordinates of the point P (x, y) which divides the line segment joining the points A (x_{1}, y_{1}) and B (x_{2}, y_{2}), internally, in the ratio m_{1} : m_{2} are

This is the **Section Formula**.

**Section Formula Examples**

Let us try to understand it better through an example:

**Example:** In what ratio does the point (– 4, 6) divide the line segment joining the points A (– 6, 10) and B (3, – 8)?

**Solution:**

The co-ordinates of point A and B are (– 6, 10) and (3, – 8).

Let (– 4, 6) divide AB internally in the ratio m_{1} : m_{2}.

Here,

x = -4 and y = 6

x_{1} = -6, x_{2} = 3,

And y_{1} = 10, y_{2} = -8

Using the section formula,

Recall that if (x, y) = (a, b) then x = a and y = b.

Now, taking

= 7m_{1} = 2m_{2}

= m_{1}: m_{2} = 2: 7

We should verify that the ratio satisfies the y-coordinate also.

Now,

Dividing throughout by m_{2},

Therefore, the point (– 4, 6) divides the line segment joining the points A (– 6, 10) and B (3, – 8) in the ratio 2: 7.

**Read More**– What is Circle – Definition, Formula, Properties, & Examples