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# Two Dimensional Coordinate Geometry

Two-dimensional coordinate geometry deals with the x and y coordinates which are represented in a coordinate plane or Cartesian plane. A coordinate plane has two axes, the one which is horizontal is known as X-axis and the one which is vertical is known as Y-axis. A point p(x,y)Â is represented in the X-Y plane, where x and y are the coordinates of the point,Â as shown below.

‘Oâ€™ is known as origin whose coordinate is (0,0).

The perpendicular distance ofÂ p(x,y) from Xâˆ’axis and Yâˆ’axis is ‘y’ and ‘x’ respectively.

Also, check: Analytic geometry

## Coordinate Geometry in Two Dimensional Plane

In coordinate geometry, first, we learn about locating the points in a Cartesian plane.

For example, the point (2,3) is 3 units away from the Xâˆ’axis measured along the positive Yâˆ’axis and 2 units away from Yâˆ’axis measured along the positive Xâˆ’axis.

The points having x-coordinate as â€˜0â€² lie on the Yâˆ’axis and points having y-coordinate as ‘0’ lie on the Xâˆ’axis.

For example, the points (2,0), (5,0) lie on Xâˆ’axis and the points (0,âˆ’3), (0,7) lie on Yâˆ’axis.

### Distance between two points

Consider two points

$$\begin{array}{l}A(x_1,y_1)\end{array}$$
and
$$\begin{array}{l}B(x_2,y_2)\end{array}$$
in an XYÂ plane.

Then the distance between A and B is,

$$\begin{array}{l}AB = \sqrt{(x_2 ~-~ x_1)^2 + (y_2~-~y_1)^2}\end{array}$$

For example; distance between the points A(2,-3)Â and B(5,1)Â is,

$$\begin{array}{l}AB = \sqrt{(5~-~2)^2 ~+~ (1-(-3))^2} = \sqrt{3^2~ + ~4^2} = \sqrt{25} = 5\end{array}$$
units

Similarly, distance between a point P(x,y)Â from the origin is,

$$\begin{array}{l}OP = \sqrt{(x-0)^2 ~+ ~(y-0)^2} = \sqrt{(x^2~ +~ y^2)}\end{array}$$

### Reflection of a point across the X-axis

Reflection of a point P(x,y) across the Xâˆ’axis is Q(x,âˆ’y), which is found by changing the sign of the y-coordinate of P(x,y).

### Reflection of a point across the Y-axis

Reflection of a point P(x, y)Â across the Y-axisÂ is Q(-x,y), which is found by changing the sign of theÂ x-coordinate of P(x,y).

### Section Formula

Consider two points

$$\begin{array}{l}A(x_1,y_1)\end{array}$$
and
$$\begin{array}{l}B(x_2,y_2)\end{array}$$
in anÂ XYÂ plane.

P(x,y)Â is a point which divides the line segment ABÂ internally in the ratio m:n,

Then, the x-coordinate of P is,

$$\begin{array}{l}x = \frac{m~x_2~ + ~n~x_1}{m~ + ~n}\end{array}$$

the

$$\begin{array}{l}y\end{array}$$
-coordinate of
$$\begin{array}{l}P\end{array}$$
is,

$$\begin{array}{l}y = \frac{m~y_2~ + ~n~y_1}{m~ + ~n}\end{array}$$

If m = n, then PÂ is the mid-point of the line segment AB. Then coordinates of the point PÂ is,

$$\begin{array}{l}y = \frac{m~y_2~ + ~n~y_1}{m~ + ~n}\end{array}$$

If the point P(x,y)Â is dividing the line segment ABÂ externally in the ratio m:n,

Then, the x-coordinate of P is,

$$\begin{array}{l}x = \frac{m~x_2~ – ~n~x_1}{m ~- ~n}\end{array}$$

And y-coordinate of PÂ is,

$$\begin{array}{l}y = \frac{m~y_2 ~- ~n~y_1}{m~ – ~n}\end{array}$$

### Solved Example

Example: Find the coordinates of the point which divides the line segment joining P(-3,-4)Â andÂ Q(6,8)Â in the ratio 1:2.

Let M(x,y)Â be the point which divides PQÂ in the ratio 1:2, then

$$\begin{array}{l}x\end{array}$$
=
$$\begin{array}{l}\frac{1 ~Ã—~ 6 ~+~ 2~ Ã—~ (-3)}{1 ~+~ 2}\end{array}$$
=
$$\begin{array}{l} \frac{6~ -~ 6}{3}\end{array}$$
=
$$\begin{array}{l}0\end{array}$$

$$\begin{array}{l}y\end{array}$$
=
$$\begin{array}{l}\frac{1 ~Ã—~ 8 ~+~ 2~ Ã—~ (-4)}{1 ~+ ~2} \end{array}$$
=
$$\begin{array}{l}\frac{8 – 8}{3}\end{array}$$
=
$$\begin{array}{l}0\end{array}$$

Therefore origin

$$\begin{array}{l}(0,0)\end{array}$$
divides the point
$$\begin{array}{l}PQ\end{array}$$
in the ratio
$$\begin{array}{l}1:2\end{array}$$
.

### Area of a triangle formed by joining three points

The area of the triangle whose vertices are

$$\begin{array}{l}(x_1,y_1 ),(x_2,y_2)\end{array}$$
and
$$\begin{array}{l}(x_3,y_3)\end{array}$$
is

$$\begin{array}{l}\frac{1}{2}|x_1 (y_2~ -~ y_3)~ + ~x_2(y_3~ – ~y_1)~ +~ x_3(y_1~ – ~y_2)|\end{array}$$

If the area of a triangle whose vertices are

$$\begin{array}{l}(x_1,y_1),(x_2,y_2)\end{array}$$
and
$$\begin{array}{l}(x_3,y_3)\end{array}$$
is zero, then the three points are collinear.

We have learnt about two-dimensional coordinate geometry until now.Â To solve more problems on topic Coordinate Geometry visit BYJU’S which provides detailed and step by step solutions to all questions in an NCERT Books. Also, take free tests to practice for exams.

#### 1 Comment

1. Samriti Sharma