Rectangular Cartesian Coordinate System

• The Cartesian System of Rectangular Coordinates, also known as The Rectangular Coordinate System or Cartesian Plane is a method of representing points in a plane using two perpendicular number lines that intersect at a common point called the origin.
• It was developed by the French mathematician René Descartes (hence "Cartesian").
• The system comprises of two axes, known as X-axis and Y-axis. X-axis is a horizontal number line, with positive orientation towards the right, and negative towards the left. Y-axis is a vertical number line, with positive orientation upward and negative downward.
• Any point \(P\) in the plane is represented by an ordered pair \((x,y)\), where
  • \(x\) : Perpendicular distance of \(P\) from \(y\)-axis, measured along \(x\)-axis. It is called abscissa, or simply x-coordinate.
  • \(y\) : Perpendicular distance of \(P\) from \(x\)-axis, measured along \(y\)-axis. It is called ordinate, or simply y-coordinate.
  • The point of intersection of two axes, known as the Origin, is denoted by \(O(0,0)\).
• The axes divide the plane into four quadrants (I,II,III,IV), numbered counterclockwise, starting from right top side.

Distance Formula

1.1 The distance between two points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) is given by: \[ PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

Proof: The proof comes directly from Pythagoras theorem applied to the right triangle formed by the horizontal and vertical separations.

Section Formula

2.1 Internal division: Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) be two points. Let \(R(x,y)\) divide the line segment \(PQ\) in the ratio \(m:n\). When \(R\) lies between \(P\) and \(Q\) (i.e., \(R\) divides \(PQ\) internally): \[ R(x,y)= \left( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n} \right) \]

Proof: Let \(P(x_1,y_1)\), \(Q(x_2,y_2)\), and \(R(x,y)\) divide \(PQ\) internally in the ratio \(m:n\).

Draw perpendiculars from \(P\), \(R\), and \(Q\) to the \(x\)-axis, meeting it at \(P'\), \(R'\), and \(Q'\).
The \(x\)-coordinates of these feet are \(x_1,x,x_2\).

By similar triangles, \[ \frac{P'R'}{R'Q'} = \frac{PR}{RQ} = \frac{m}{n}. \]

Since \(P'R'=x-x_1\) and \(R'Q'=x_2-x\), \[ \frac{x-x_1}{x_2-x} = \frac{m}{n} \] \[ n(x-x_1)=m(x_2-x) \] \[ nx-nx_1=mx_2-mx \] \[ nx+mx=mx_2+nx_1 \] \[ x(m+n)=mx_2+nx_1 \] \[ \boxed{ x=\frac{mx_2+nx_1}{m+n} } \]

Similarly, drawing perpendiculars to the \(y\)-axis gives: \[ \boxed{ y=\frac{my_2+ny_1}{m+n} } \]

2.2 External division: Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) be two points. Let \(R(x,y)\) divide the line segment \(PQ\) in the ratio \(m:n\). When \(R\) lies outside the segment \(PQ\), on the line \(PQ\) extended: \[ R(x,y)= \left( \frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n} \right) \]

Proof: On similar lines as above proof, but \(R\) lies outside \(PQ\). Here \(R'\) lies outside \(P'Q'\), so: \[ \frac{x-x_1}{x-x_2} = \frac{m}{n} \] \[ n(x-x_1)=m(x-x_2) \] \[ nx-nx_1=mx-mx_2 \] \[ nx-mx=nx_1-mx_2 \] \[ x(n-m)=nx_1-mx_2 \] \[ \boxed{ x=\frac{mx_2-nx_1}{m-n} } \]

Similarly: \[ \boxed{ y=\frac{my_2-ny_1}{m-n} } \]

Exercises

1. If the equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) represents a pair of straight lines, show that they intersect at the point \[ \left( \frac{hf-bg}{ab-h^2}, \frac{hg-af}{ab-h^2} \right). \]
2. Show that the pair of straight lines through origin, and perpendicular to the lines \(ax^2 + 2hxy + by^2 = 0\) are given by \(bx^2 - 2hxy + ay^2 = 0\).
3. Show that the four lines given by \(2x^2 + 3xy - 2y^2 = 0\) and \(2x^2 + 3xy - 2y^2 - 17x + 6y + 8 = 0\) form a rectangle. Also,

   (a) Find the area of the rectangle.

   (b) Find the joint equation of the diagonals.

   (c) Find the point of intersection of the diagonals.

   (d) Find the angle between the diagonals.

   (e) Find the equation of the circle circumscribing this rectangle.