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) 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.

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(x1, y1) and B(x2, y2) in an XY plane.

Then the distance between A and B is,

$\begin{array}{l}��=\sqrt{({�}_2-{�}_1{)}^2+({�}_2-{�}_1{)}^2}\end{array}$

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

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

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

$\begin{array}{l}��=\sqrt{(�-0{)}^2+(�-0{)}^2}=\sqrt{({�}^2+{�}^2)}\end{array}$

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(x1, y1) and B(x2, y2) 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,

$\begin{array}{l}�=\frac{�{�}_2+�{�}_1}{�+�}\end{array}$

And, the y-coordinate of P is,

$\begin{array}{l}�=\frac{�{�}_2+�{�}_1}{�+�}\end{array}$

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

$\begin{array}{l}�=\frac{�{�}_2+�{�}_1}{�+�}\end{array}$

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,

$\begin{array}{l}�=\frac{�{�}_2–�{�}_1}{�-�}\end{array}$

And y-coordinate of P is,

$\begin{array}{l}�=\frac{�{�}_2-�{�}_1}{�–�}\end{array}$

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

$\begin{array}{l}�=\frac{1\times 6+2\times (-3)}{1+2}=\frac{6-6}{3}=0\end{array}$

$\begin{array}{l}�=\frac{1\times 8+2\times (-4)}{1+2}=\frac{8–8}{3}=0\end{array}$

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 (x1, y1), (x2, y2) and (x3, y3) is

$\begin{array}{l}\frac{1}{2}|{�}_1({�}_2-{�}_3)+{�}_2({�}_3–{�}_1)+{�}_3({�}_1–{�}_2)|\end{array}$

If the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is zero, then the three points are collinear.

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