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) is represented in the X-Y plane, where x and y are the coordinates of the point, as shown below.

‘O’ is known as origin whose coordinate is (0,0).

The perpendicular distance of p(x,y) from X−axis and Y−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 X−axis measured along the positive Y−axis and 2 units away from Y−axis measured along the positive X−axis.

The points having x-coordinate as ‘0′ lie on the Y−axis and points having y-coordinate as ‘0’ lie on the X−axis.

For example, the points (2,0), (5,0) lie on X−axis and the points (0,−3), (0,7) lie on Y−axis.

Distance between two points

Consider two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in an XY plane.

Then the distance between A and B is,

\(AB = \sqrt{(x_2 ~-~ x_1)^2 + (y_2~-~y_1)^2}\)

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

\(AB = \sqrt{(5~-~2)^2 ~+~ (1-(-3))^2} = \sqrt{3^2~ + ~4^2} = \sqrt{25} = 5\) units

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

\(OP = \sqrt{(x-0)^2 ~+ ~(y-0)^2} = \sqrt{(x^2~ +~ y^2)}\)

Reflection of a point across the X-axis

Reflection of a point P(x,y) across the X−axis is Q(x,−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) across the Y-axis is Q(-x,y), which is found by changing the sign of the x-coordinate of P(x,y).

Section Formula

Consider two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in an XY plane.

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

Then, the x-coordinate of P is,

\(x = \frac{m~x_2~ + ~n~x_1}{m~ + ~n}\)

the \(y\)-coordinate of \(P\) is,

\(y = \frac{m~y_2~ + ~n~y_1}{m~ + ~n}\)

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

\(y = \frac{m~y_2~ + ~n~y_1}{m~ + ~n}\)

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

Then, the x-coordinate of P is,

\(x = \frac{m~x_2~ – ~n~x_1}{m ~- ~n}\)

And y-coordinate of P is,

\(y = \frac{m~y_2 ~- ~n~y_1}{m~ – ~n}\)

Solved Example

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

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

\(x\) = \(\frac{1 ~×~ 6 ~+~ 2~ ×~ (-3)}{1 ~+~ 2}\) = \( \frac{6~ -~ 6}{3}\) = \(0\)

\(y\) = \(\frac{1 ~×~ 8 ~+~ 2~ ×~ (-4)}{1 ~+ ~2} \) = \(\frac{8 – 8}{3}\) = \(0\)

Therefore origin \((0,0)\) divides the point \(PQ\) in the ratio \(1:2\).

Area of a triangle formed by joining three points

The area of the triangle whose vertices are \((x_1,y_1 ),(x_2,y_2)\) and \((x_3,y_3)\) is

\(\frac{1}{2}|x_1 (y_2~ -~ y_3)~ + ~x_2(y_3~ – ~y_1)~ +~ x_3(y_1~ – ~y_2)|\)

If the area of a triangle whose vertices are \((x_1,y_1),(x_2,y_2)\) and \((x_3,y_3)\) is zero, then the three points are collinear.

We have learnt about two-dimensional coordinate geometry until now. 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.