**RECTANGULAR CO-ORDINATE AXES:** Let X’OX and Y’OY be two mutually perpendicular lines through any point O in the plane of the paper. We call the point O, the origin. Now choose a convenient unit of length and starting from the origin as zero, mark off a number scale on the horizontal line X’OX, positive to the right of the origin O and negative to the left of origin O. Also, mark off the same scale on the vertical line Y’OY, positive upwards and negative downwards of the origin O. The line X’OX is called the x-axis or axis of x, the line Y’OY is known as the y-axis or axis of y, and the two lines taken together are called the co-ordinate axes or the axes of co-ordinates.

**CARTESIAN CO-ORDINATES OF A POINT: **Let X’OX and Y’OY be the coordinate axes, and let P be any point in the plane. Draw perpendiculars PM and PN from P on x and y-axis respectively. The length of the directed line segment OM in the units of scale chosen is called the x-coordinate or abscissa of point P. Similarly, the length of the directed line segment ON on the same scale is called the y-coordinate or ordinate of point P. Let OM = x and ON = y. Then the position of the point P in the plane with respect to the coordinate axes is represented by the ordered pair (x, y). The ordered pair (x, y) is called the coordinates of point P.Thus, for a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively.

i. Quadrant: x > o, y > o

ii. Quadrant: x < 0, y > 0

iii. Quadrant: x < 0, y < 0

iv. Quadrant: x > 0, y < 0

The coordinates of the origin are taken as (0, 0). The coordinates of any point on x-axis are of the form (x, 0) and the coordinates of any point on y-axis are of the form (0, y). Thus, if the abscissa of a point is zero, if would lie somewhere on the y-axis and if its ordinate is zero it would lie on x-axis.

**DISTANCE BETWEEN TWO POINTS:** The Distance between any two points in the plane is the length of the line segment joining them.

The distance between the points P (x₁, y₁) and Q (x₂, y₂) is given by \(PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\)

i.e., \(PQ=\sqrt{{{\left( difference\,of\,abscissa \right)}^{2}}+{{\left( difference\,of\,ordinates \right)}^{2}}}\)

If Q is the origin and P(x, y) is any point, then from the above formula,

We have \(OP=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{y}^{2}}}\)

**Example: 1**: If the points (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

**Solution: **Let P (x, y), Q (a + b, b – a) and R (a – b, a + b) be the given points.

Then, PQ = QR [given]

⇒ \(\sqrt{{{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( b-a \right) \right)}^{2}}}=\sqrt{{{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}}\)

⇒ [x – (a + b)²] + [y – (b – a)²] = [x – (a – b)²] + [y – (a + b)²]

⇒ x² – 2x(a + b) + (a + b)² + y² – 2y (b – a) + (b – a)²

= x² + (a – b)² – 2x(a – b) + y² – 2y (a + b) + (a + b)²

⇒ – 2x (a + b) – 2y (b – a) = – 2x (a – b) – 2y (a + b)

⇒ ax + bx + by – ay = ax – bx + ay + by

⇒ 2bx = 2ay

⇒ bx = ay.

**AREA OF A TRIANGLE:** The area of a triangle, the coordinates of whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

= ½ [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)].

\(\Delta =\frac{1}{2}\left| \begin{matrix}{{x}_{1}} & {{y}_{1}} & 1 \\{{x}_{2}} & {{y}_{2}} & 1 \\{{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right|\).

**CONDITION OF COLLINEARITY OF THREE POINTS: **Three points A (x₁, y₁) B (x₂, y₂) and C (x₃, y₃) are collinear.

If Area of ΔABC = 0

i.e., \(\frac{1}{2}\left| \begin{matrix}{{x}_{1}} & {{y}_{1}} & 1 \\{{x}_{2}} & {{y}_{2}} & 1 \\{{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right|=0\),

\(\left| \begin{matrix}{{x}_{1}} & {{y}_{1}} & 1 \\{{x}_{2}} & {{y}_{2}} & 1 \\{{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right|=0\).

AB + BC = AC or AC + BC = AB or AC + AB = BC

**Example 1**: If the vertices of a triangle have integral coordinates, prove that the triangle cannot be equilateral.

**Solution:** Let A(x₁, y₁), B(x_{2}, y_{2}) and C (x₃, y₃) be the vertices of a triangle ABC, where x_{i}, y_{i}; i = 1, 2, 3 are integers.

Therefore, the expression x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) is an integer and so A is a rational number.

If possible, let the triangle ABC be an equilateral triangle, then its area

Δ = \(\frac{\sqrt{3}}{4}\) (side)²

Δ = \(\frac{\sqrt{3}}{4}\) (AB)²

= \(\frac{\sqrt{3}}{4}\) (a positive integer)

= An irrational number

This is a contradiction to the fact that the area is a rational number.

Hence, the triangle cannot be equilateral.

**Example 2: **The coordinates of A, B, C are (6, 3), (-3, 5) and (4, -2) respectively and p is any point (x, y). Show that the ratio of the area of triangle PBC and ABC is \(\left| \frac{x+y-2}{7} \right|\).

**Solution: **We have,

\(\frac{Area\,of\,\Delta PBC}{Area\,of\,\Delta ABC}=\frac{\frac{1}{2}\left| x\left( 5+2 \right)+\left( -3 \right)\left( -2-y \right)+4\left( y-5 \right) \right|}{\frac{1}{2}\left| 6\left( 5+2 \right)+\left( -3 \right)\left( -2-3 \right)+4\left( 3-5 \right) \right|}\),

= \(\frac{\left| 7x+7y-14 \right|}{\left| 42+15-8 \right|}\),

= \(\frac{7\left| x+y-2 \right|}{49}\),

= \(\left| \frac{x+y-2}{7} \right|\).