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.

Property	Description
One-to-one correspondence	Every point \(\leftrightarrow\) exactly one ordered pair \((x,y)\)
Symmetry about \(x\)-axis	\((x,y)\rightarrow(x,-y)\)
Symmetry about \(y\)-axis	\((x,y)\rightarrow(-x,y)\)
Symmetry about origin	\((x,y)\rightarrow(-x,-y)\)
Symmetry about \(y=x\)	\((x,y)\rightarrow(y,x)\)

Transformation from other Systems

• From Polar to Cartesian: \[ x=r\cos\theta, \qquad y=r\sin\theta \]
• From Cartesian to Polar: \[ r=\sqrt{x^2+y^2}, \qquad \theta=\tan^{-1}\left(\frac{y}{x}\right) \]

Transformation of Axes

• Transformation of axes is very useful to simplify the equation of a curve by changing the coordinate system.
• The key idea is that the geometric point or curve does not change; only the coordinate system used to describe it changes.
• There are two fundamental transformations:
   ✓ Translation of Axes
   ✓ Rotation of Axes
  
• Translation of axes is a transformation in which the origin is shifted to a new point \((h,k)\), without rotating the axes. The new axes remain parallel to the original axes, and the scale and direction remain unchanged. The point \(P\) stays fixed in the plane, but the coordinates of \(P\) change from \((x,y)\) to \((X,Y)\).
• Rotation of axes is a transformation in which the coordinate axes are rotated about the origin by a certain angle \(\theta\), without changing the origin and without changing the scale. The point \(P\) stays fixed in the plane, the axes rotate, and the coordinates of \(P\) change from \((x,y)\) to \((X,Y)\).
• What each transformation does?

  Transformation	What changes	Why we do?
  Translation	Origin	To remove linear terms.
  Rotation	Direction of Axes	To remove \(xy\) term.
  Translation + rotation	Origin and direction	Reduce eqn. to simplest form.

  For a general conic \(ax^2+2hxy+by^2+2gx+2fy+c=0\), you can think of the process as \[ \boxed { \text{Given eqn.} \rightarrow \text{rotate to remove xy term} \rightarrow \text {translate to remove linear term.}} \]
• Invariants under translation (quantities that remain unchanged when origin is shifted):
   ✓ The coefficients of \(x^2,y^2,xy\): \(\quad a,b,h\)
   ✓ Discriminant of quadratic part: \(\quad ab-h^2\)
• Invariants under rotation (quantities that remain unchanged when axes are rotated)
   ✓ The coefficient sum: \(\quad a+b\)
   ✓ Discriminant of quadratic part: \(\quad ab-h^2\)
   ✓ Distance from origin \((x^2+y^2 = X^2 + Y^2)\)

1.1 Consider translation of axes in which the origin is shifted to a new point \((h,k)\), and the coordinates of \(P\) change from \((x,y)\) to \((X,Y)\). Prove the translation formulas: \[ \boxed { X = x - h, \quad Y = y - k }\] Or Equivalently, \[ \boxed { x = X + h, \quad y = Y + k } \]

Proof: Let the original axes be \(OX\) and \(OY\) (x-axis and y-axis).
Let the new origin be \(O'(h,k)\). Let the new axes be \(O'X'\) and \(O'Y'\), parallel to the original axes. Let a point \(P\) have

• Old coordinates \((x,y)\) w.r.t. OXY.
• new coordinates \((X,Y)\) w.r.t. O'X'Y'.

1.2 Consider rotation of axes in which the coordinate axes are rotated by an angle \(\theta\) about the origin, and the coordinates of \(P\) change from \((x,y)\) to \((X,Y)\). Prove the rotation formulas: \[ \boxed { x = X\cos\theta - Y\sin\theta } \qquad \boxed { y = X\sin\theta + Y\cos\theta } \] and the inverse transformation is, \[ \boxed { X = x\cos\theta + y\sin\theta } \qquad \boxed { Y = -x\sin\theta + y\cos\theta } \]

Proof: Let the original axes be \(OX\) and \(OY\) (x-axis and y-axis).
Let the new axes be \(OX'\) and \(OY'\), obtained by rotating the original axes counterclockwise by angle \(\theta\). Let a point \(P\) have

• Old coordinates \((x,y)\) w.r.t. OXY.
• new coordinates \((X,Y)\) w.r.t. OX'Y'.

• This transformation can also be written as \[ \begin{pmatrix} X\\ Y \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} \]
• The rotation matrix, \(R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} \) is orthogonal, i.e., \(R^{-1} = R^T\). Additionally, det(\(R\)) = 1.
• The most general transformation of axes consists of translation (to \((h,k)\)), followed by rotation (of angle \(\theta\) counterclockwise). \[ x = h + X\cos\theta - Y\sin\theta \] \[ y = k + X\sin\theta + Y\cos\theta \]

1.3 Suppose a given conic equation is \(ax^2+2hxy+by^2+2gx+2fy+c=0\) in original coordinate axes. If coordinate axes are rotated by an angle \(\theta\) such that the xy term is removed after rotation, then show that \(\tan2\theta = \frac{2h}{a-b}\).

Proof: Substitute \(x = X\cos\theta - Y\sin\theta\) and \(y = X\sin\theta + Y\cos\theta\) into the equation. The coefficient of \(XY\) in the transformed equation shall be \[ 2(b-a)\sin\theta\cos\theta + 2h(\cos^2\theta - \sin^2\theta) \] Equating it to \(0\) gives \[ (b-a)\sin2\theta +2h\cos2\theta = 0 \] \[ \boxed{\tan2\theta = \frac{2h}{a-b}} \]

• Note: If \(a=b\) in the given conic equation, a rotation of \( 45^\circ\) (or \(\pi/4\)) is required to remove the \(xy\) term.
• Note: There are two values of \(2\theta\) in \([0,2\pi)\) differing by \(\pi\), giving two values of \(\theta\) differing by \(\pi/2\). Either choice will eliminate the \(xy\) term (they correspond to swapping the new \(x\) and \(y\) axes).

Distance Formula

2.1 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

3.1 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: \[ \boxed{ y=\frac{my_2+ny_1}{m+n} } \]

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

• The coordinates of the mid-point of the line segment joining \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) is given by \(\left(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2}\right)\).
• Suppose \(A\) and \(B\) are two fixed points, and \(P\) divides \(AB\) internally in the ratio \(m:n\), and \(Q\) divides \(AB\) externally in the ratio \(m:n\), then \(P\) and \(Q\) are called harmonic conjugates w.r.t. \(A\) and \(B\).
  The key property is \(\frac{AP}{PB} = -\frac{AQ}{QB}\), where lengths are directed lengths. From this property, one can derive \(\frac{1}{AP} + \frac{1}{AQ} = \frac{2}{AB}\).

Key Triangle Points

• Euler line
  • The Euler line is one of the most beautiful structural results in triangle geometry.
  • For a triangle \(ABC\), the Euler Line is the line passing through its Centroid (\(G\)), Circumcentre (\(O\)) and Orthocentre (\(H\)). Thus, O, G, H are collinear.
  • For an equilateral triangle, all three points coincide, so the Euler line is not uniquely defined.
  • The centroid divides the segment joining the circumcentre and orthocentre in the ratio \[ \boxed{ OG:GH = 1:2 } \quad \text{ or equivalently } \quad \boxed { OG = \frac{1}{2}GH } \]

Area of Triangle

4.1 If the coordinates of the vertices of a triangle are \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\), then its area is \[ \Delta= \frac{1}{2} \left| x_1(y_2-y_3) +x_2(y_3-y_1) +x_3(y_1-y_2) \right| \] Equivalently, the area can be written in determinant form as \[ \Delta= \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

Proof: Draw perpendiculars from \(A\), \(B\), \(C\) to the \(x\)-axis, meeting it at \(A'\), \(B'\), \(C'\).
So, we have \(A'=(x_1,0)\), \(B'=(x_2,0)\) and \(C'=(x_3,0)\).
The area of triangle \(ABC\) can be expressed using trapezoids: \[ \Delta = \operatorname{Area}(ABB'A') + \operatorname{Area}(BCC'B') - \operatorname{Area}(ACC'A') \] Assuming the vertices are ordered so that \(x_1 < x_2 < x_3\) along the \(x\)-axis; otherwise signed areas handle it automatically.
Area of a trapezoid \(=\frac{1}{2}\)(sum of parallel sides)\(\times\)height.
Trapezoid \(ABB'A'\): Parallel sides \(AA'=y_1\), \(BB'=y_2\), height \(=x_2-x_1\): \[ \operatorname{Area}(ABB'A') = \frac{1}{2}(y_1+y_2)(x_2-x_1) \]
Trapezoid \(BCC'B'\): Parallel sides \(BB'=y_2\), \(CC'=y_3\), height \(=x_3-x_2\): \[ \operatorname{Area}(BCC'B') = \frac{1}{2}(y_2+y_3)(x_3-x_2) \]
Trapezoid \(ACC'A'\): Parallel sides \(AA'=y_1\), \(CC'=y_3\), height \(=x_3-x_1\): \[ \operatorname{Area}(ACC'A') = \frac{1}{2}(y_1+y_3)(x_3-x_1) \] Subtract: \[ \Delta = \frac{1}{2}(y_1+y_2)(x_2-x_1) + \frac{1}{2}(y_2+y_3)(x_3-x_2) - \frac{1}{2}(y_1+y_3)(x_3-x_1) \] Simplifying gives \[ \boxed{ \Delta=\frac{1}{2} \left[ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right] } \]

• Alternative forms of area of triangle:
  • \(\Delta\ = \frac{1}{2}bh \quad\) if base: \(b\) and height: \(h\).
  • \(\Delta\ = \frac{1}{2}ab SinC \quad\) if sides \(a\), \(b\) and included angle \(C\) is given.
  • \(\Delta\ = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s=\frac{a+b+c}{2}\) and \(a\), \(b\), \(c\) are sides. This is well known Heron's Formula.
  • \(\Delta\ = \frac{\sqrt{3}}{4}a^2 \) for an equilateral triangle with side \(a\).
  • \(\Delta\ = \frac{abc}{4R} \) for a triangle with circumradius R.
  • \(\Delta\ = rs \) for a triangle with inradius R.
• Three points \(A\), \(B\) and \(C\) are collinear if and only if the area of the triangle is zero, i.e. \( \Delta = 0\).
  \( \boxed{\text{Collinear}\iff \Delta=0} \)
• The area of a quadrilateral having vertices \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\), \(D(x_4,y_4)\) in order is given by \[ \text{Area} = \frac{1}{2} \left| (x_1y_2-x_2y_1) +(x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) +(x_4y_1-x_1y_4) \right| \]
• Area of polygon having vertices \((x_1,y_1)\), \((x_2,y_2)\), ... \((x_n, y_n)\) in order is given by \[ \text{Area}=\frac{1}{2} \left| \sum_i \left(x_i y_{i+1}-x_{i+1}y_i\right) \right| \]

Exercises

1. A point has coordinates \( (3,4) \) in the original system.

   (a) Find the coordinates after rotating the axes by \( 45^\circ\) counterclockwise.

   (b) In the original coordinate system, find the coordinates of this point \( (3,4) \) after rotating the point itself by \( 45^\circ\).

   Note that rotating the axes is not the same as rotating the point.

2. Transform the equation \( x^2+ y^2 - 6x -8y +9 = 0 \) by shifting the Origin to \((3,4)\).
   Note that the translation has removed the linear terms.
3. Transform the equation \(2x^2+ y^2 - 12x + 4y + 14 = 0\) by translating the axes to remove the first-degree terms.
4. Transform the equation \( 5x^2+6xy+5y^2=8 \) by rotating the axes to eliminate the \( xy \) term.
5. Transform the equation \( x^2+8xy+y^2=16 \) by rotating the axes to eliminate the \( xy \) term.
6. Prove that the distance from origin remains unchanged if we rotate the coordinate axes by angle \(\theta\) counterclockwise.