Cartesian Plane and Coordinates

The Cartesian plane is formed by two perpendicular number lines intersecting at the origin (0,0). The horizontal line is called the x-axis and the vertical line is called the y-axis. These axes divide the plane into four quadrants numbered I, II, III, and IV in an anti-clockwise direction starting from the positive x-axis.

Every point on the Cartesian plane is represented by an ordered pair \((x, y)\), where \(x\) is the abscissa (distance from the y-axis) and \(y\) is the ordinate (distance from the x-axis). The coordinates indicate the position of the point relative to the origin.

Key Properties

• The coordinate of a point on the x-axis is of the form \((x, 0)\).
• The coordinate of a point on the y-axis is of the form \((0, y)\).
• The sign of the coordinates depends on the quadrant:

Worked Illustration

Plot the point \(P(3, 4)\) on the Cartesian plane.

Solution:

1. Start at the origin \((0,0)\).
2. Move 3 units along the positive x-axis.
3. From there, move 4 units up along the positive y-axis.
4. Mark the point \(P\) at this position.

Practice Set

Level 1 – Easy

• Identify the coordinates of the origin.
• State the coordinates of a point on the x-axis 5 units to the right of the origin.
• State the coordinates of a point on the y-axis 7 units above the origin.

Level 2 – Moderate

• Determine the quadrant of the point \((-4, 6)\).
• Plot the points \((−3, −2)\) and \((5, −1)\) and state their quadrants.

Level 3 – Challenging

• Given a point \(Q(x, y)\) lies in the second quadrant and \(x + y = 0\), find the possible coordinates of \(Q\).

Answer Key

• Origin coordinates: \((0,0)\).
• Point on x-axis 5 units right: \((5,0)\).
• Point on y-axis 7 units above: \((0,7)\).
• Quadrant of \((-4,6)\): II.
• Point \((-3,-2)\) lies in III; point \((5,-1)\) lies in IV.
• For \(x + y = 0\) in quadrant II, \(x < 0\) and \(y > 0\), so \(Q = (-a, a)\) where \(a > 0\).

Quick Reference

• Origin: \((0,0)\)
• Quadrants: I (+,+), II (−,+), III (−,−), IV (+,−)
• Coordinates: \((x,y)\) where \(x\) is abscissa and \(y\) is ordinate

Glossary

• Abscissa: The x-coordinate of a point.
• Ordinate: The y-coordinate of a point.
• Origin: The point \((0,0)\) where the axes intersect.
• Quadrant: One of the four regions into which the plane is divided by the axes.

Distance Formula

The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) in the Cartesian plane is the length of the line segment \(PQ\). It is calculated using the Pythagorean theorem.

Formula Derivation

Consider the points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\). Construct a right triangle by drawing perpendiculars from these points to the axes.

The horizontal side length is \(|x_2 - x_1|\) and the vertical side length is \(|y_2 - y_1|\).

By the Pythagorean theorem, the distance \(d\) is:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Worked Illustration

Find the distance between points \(A(3, 2)\) and \(B(7, 6)\).

Solution:

Using the distance formula,

\[ AB = \sqrt{(7 - 3)^2 + (6 - 2)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

Practice Set

Level 1 – Easy

• Calculate the distance between \((0,0)\) and \((3,4)\).
• Find the distance between \((1,2)\) and \((1,5)\).

Level 2 – Moderate

• Find the distance between \((-2, -3)\) and \((4, 1)\).
• Calculate the distance of the point \((5, 12)\) from the origin.

Level 3 – Challenging

• Prove that the points \((1,2)\), \((4,6)\), and \((7,10)\) are collinear using the distance formula.

Answer Key

• Distance between \((0,0)\) and \((3,4)\) is 5 units.
• Distance between \((1,2)\) and \((1,5)\) is 3 units.
• Distance between \((-2,-3)\) and \((4,1)\) is \(\sqrt{(4+2)^2 + (1+3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\).
• Distance of \((5,12)\) from origin is \(\sqrt{5^2 + 12^2} = 13\).
• Check collinearity: \(AB = \sqrt{(4-1)^2 + (6-2)^2} = 5\), \(BC = \sqrt{(7-4)^2 + (10-6)^2} = 5\), \(AC = \sqrt{(7-1)^2 + (10-2)^2} = 10\). Since \(AB + BC = AC\), points are collinear.

Quick Reference

• Distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
• Distance from origin: \(\sqrt{x^2 + y^2}\)

Glossary

• Distance: The length between two points in the plane.
• Collinear points: Points lying on the same straight line.

Midpoint Formula

The midpoint of a line segment joining two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is the point \(M\) that divides the segment into two equal parts.

Formula Derivation

The coordinates of the midpoint \(M\) are the averages of the corresponding coordinates of \(A\) and \(B\):

\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Worked Illustration

Find the midpoint of the segment joining \(P(2, 3)\) and \(Q(6, 7)\).

Solution:

\[ M = \left( \frac{2 + 6}{2}, \frac{3 + 7}{2} \right) = (4, 5) \]

Practice Set

Level 1 – Easy

• Find the midpoint of \((0,0)\) and \((4,4)\).
• Find the midpoint of \((1,5)\) and \((3,9)\).

Level 2 – Moderate

• Find the midpoint of \((-2, 7)\) and \((6, -1)\).
• Find the midpoint of \((x, y)\) and \((2x, 3y)\).

Level 3 – Challenging

• Given the midpoint \(M(3, 4)\) of segment \(AB\), and point \(A(1, 2)\), find the coordinates of \(B\).

Answer Key

• Midpoint of \((0,0)\) and \((4,4)\) is \((2,2)\).
• Midpoint of \((1,5)\) and \((3,9)\) is \((2,7)\).
• Midpoint of \((-2,7)\) and \((6,-1)\) is \((2,3)\).
• Midpoint of \((x,y)\) and \((2x,3y)\) is \(\left( \frac{3x}{2}, 2y \right)\).
• Coordinates of \(B\) are \((5,6)\) since \(3 = \frac{1 + x_B}{2}\) and \(4 = \frac{2 + y_B}{2}\).

Quick Reference

• Midpoint formula: \(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

Glossary

• Midpoint: The point dividing a line segment into two equal parts.

Section Formula

The section formula gives the coordinates of a point \(P\) dividing the line segment joining two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m:n\).

Formula Derivation

If \(P\) divides \(AB\) internally in the ratio \(m:n\), then the coordinates of \(P\) are:

\[ x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n} \]

Worked Illustration

Find the coordinates of the point dividing the segment joining \(A(2, 3)\) and \(B(8, 7)\) in the ratio \(3:2\).

Solution:

\[ x = \frac{3 \times 8 + 2 \times 2}{3 + 2} = \frac{24 + 4}{5} = \frac{28}{5} = 5.6 \]

\[ y = \frac{3 \times 7 + 2 \times 3}{3 + 2} = \frac{21 + 6}{5} = \frac{27}{5} = 5.4 \]

Coordinates of the point are \((5.6, 5.4)\).

Practice Set

Level 1 – Easy

• Find the point dividing \((0,0)\) and \((4,4)\) in the ratio \(1:1\).
• Find the point dividing \((1,2)\) and \((3,6)\) in the ratio \(2:3\).

Level 2 – Moderate

• Find the point dividing \((-2, 5)\) and \((4, -1)\) in the ratio \(3:1\).
• Find the ratio in which \((2, 3)\) divides the segment joining \((1, 1)\) and \((5, 7)\).

Level 3 – Challenging

• Find the coordinates of the point dividing the segment joining \(A(-6, 10)\) and \(B(3, -8)\) in the ratio \(2:7\).

Answer Key

• Point dividing \((0,0)\) and \((4,4)\) in ratio \(1:1\) is \((2,2)\).
• Point dividing \((1,2)\) and \((3,6)\) in ratio \(2:3\) is \(\left( \frac{2 \times 3 + 3 \times 1}{5}, \frac{2 \times 6 + 3 \times 2}{5} \right) = (1.8, 3.6)\).
• Point dividing \((-2,5)\) and \((4,-1)\) in ratio \(3:1\) is \(\left( \frac{3 \times 4 + 1 \times (-2)}{4}, \frac{3 \times (-1) + 1 \times 5}{4} \right) = (2.5, 0.5)\).
• Ratio in which \((2,3)\) divides \((1,1)\) and \((5,7)\) is \(1:2\).
• Coordinates dividing \(A(-6,10)\) and \(B(3,-8)\) in ratio \(2:7\) are \((-4,6)\).

Quick Reference

• Section formula (internal division): \(\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)\)

Glossary

• Section formula: Formula to find a point dividing a segment in a given ratio.
• Internal division: When the point lies between the two points.