Cartesian Rectangular Coordinate System in a Plane

Introduction to the Cartesian Coordinate System

The Cartesian coordinate system, named after the French mathematician René Descartes, is a fundamental tool in mathematics for describing the position of points in a plane or in space. It provides a systematic way to represent geometric shapes using algebraic equations. This system is built upon two perpendicular lines, called axes, which intersect at a point known as the origin.

The Axes

In a plane, the Cartesian system uses two axes:

• The x-axis: This is a horizontal line. It is also referred to as the abscissa.
• The y-axis: This is a vertical line, perpendicular to the x-axis. It is also referred to as the ordinate.

These two axes divide the plane into four regions called quadrants.

The Origin

The point where the x-axis and y-axis intersect is called the origin. It is denoted by the symbol 'O'. The coordinates of the origin are (0, 0). All measurements of distance along the axes start from the origin.

Coordinates of a Point

Any point P in the Cartesian plane can be uniquely identified by a pair of numbers called its coordinates. These coordinates are written as an ordered pair (x, y), where:

• x is the x-coordinate (abscissa): It represents the perpendicular distance of the point from the y-axis. If x is positive, the point is to the right of the y-axis; if x is negative, it is to the left.
• y is the y-coordinate (ordinate): It represents the perpendicular distance of the point from the x-axis. If y is positive, the point is above the x-axis; if y is negative, it is below.

The order of the coordinates is crucial. The point (3, 5) is different from the point (5, 3).

Quadrants

The x-axis and y-axis divide the plane into four quadrants, numbered counterclockwise starting from the upper right:

Quadrant	x-coordinate (Abscissa)	y-coordinate (Ordinate)	Sign of Coordinates
Quadrant I (Q1)	Positive	Positive	(+, +)
Quadrant II (Q2)	Negative	Positive	(-, +)
Quadrant III (Q3)	Negative	Negative	(-, -)
Quadrant IV (Q4)	Positive	Negative	(+, -)

Points lying on the axes do not belong to any quadrant. For example, a point on the x-axis has a y-coordinate of 0, and a point on the y-axis has an x-coordinate of 0.

Plotting Points

To plot a point with coordinates (x, y):

1. Start at the origin (0, 0).
2. Move horizontally along the x-axis by 'x' units. Move to the right if 'x' is positive, and to the left if 'x' is negative.
3. From that position, move vertically parallel to the y-axis by 'y' units. Move upwards if 'y' is positive, and downwards if 'y' is negative.
4. The point where you end up is the location of (x, y).

Example: To plot the point (-2, 3): Start at the origin, move 2 units to the left (along the negative x-axis), then move 3 units up (parallel to the positive y-axis). This point lies in Quadrant II.

Distance Formula

The distance between two points P1(x1, y1) and P2(x2, y2) in the Cartesian plane can be calculated using the distance formula, which is derived from the Pythagorean theorem.

Consider the two points P1(x1, y1) and P2(x2, y2). Draw a line segment connecting them. Construct a right-angled triangle with P1 and P2 as two vertices, and the third vertex Q having coordinates (x2, y1) or (x1, y2). The lengths of the two perpendicular sides of this triangle are |x2 - x1| and |y2 - y1|.

According to the Pythagorean theorem (a² + b² = c²), the square of the distance (d) between P1 and P2 is:

d² = (x2 - x1)² + (y2 - y1)²

Therefore, the distance 'd' is:

d = √((x2 - x1)² + (y2 - y1)²)

Example: Find the distance between points A(2, 3) and B(5, 7).

Here, x1 = 2, y1 = 3, x2 = 5, y2 = 7.

d = √((5 - 2)² + (7 - 3)²) d = √(3² + 4²) d = √(9 + 16) d = √(25) d = 5 units.

Section Formula

The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. Let P(x, y) be a point that divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m:n.

The coordinates of P(x, y) are given by:

x = (m*x2 + n*x1) / (m + n)

y = (m*y2 + n*y1) / (m + n)

This can be written concisely as:

P(x, y) = ( (m*x2 + n*x1) / (m + n), (m*y2 + n*y1) / (m + n) )

Special Case: Midpoint Formula

When a point P divides the line segment AB in the ratio 1:1 (i.e., it is the midpoint), the section formula simplifies to the midpoint formula. Here, m = 1 and n = 1.

x = (1*x2 + 1*x1) / (1 + 1) = (x1 + x2) / 2

y = (1*y2 + 1*y1) / (1 + 1) = (y1 + y2) / 2

So, the midpoint M of a line segment joining A(x1, y1) and B(x2, y2) is:

M(x, y) = ( (x1 + x2) / 2, (y1 + y2) / 2 )

Example: Find the coordinates of the point that divides the line segment joining (1, 2) and (4, 7) in the ratio 2:3 internally.

Here, A(x1, y1) = (1, 2), B(x2, y2) = (4, 7), m = 2, n = 3.

x = (2*4 + 3*1) / (2 + 3) = (8 + 3) / 5 = 11 / 5

y = (2*7 + 3*2) / (2 + 3) = (14 + 6) / 5 = 20 / 5 = 4

The coordinates of the point are (11/5, 4).

Example (Midpoint): Find the midpoint of the line segment joining P(3, 5) and Q(7, 9).

x = (3 + 7) / 2 = 10 / 2 = 5

y = (5 + 9) / 2 = 14 / 2 = 7

The midpoint is (5, 7).

Area of a Triangle

The area of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) can be calculated using the following formula:

Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

The absolute value is taken because area must be positive.

This formula can also be expressed using a determinant:

Area = 1/2 | &begin{vmatrix} x1 & y1 & 1 \\ x2 & y2 & 1 \\ x3 & y3 & 1 \end{vmatrix} |

Condition for Collinearity: If three points are collinear (lie on the same straight line), the area of the triangle formed by them is zero. Therefore, for points A, B, and C to be collinear, the condition is:

x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0

Example: Find the area of the triangle with vertices P(1, 2), Q(4, 5), and R(7, 8).

x1 = 1, y1 = 2 x2 = 4, y2 = 5 x3 = 7, y3 = 8

Area = 1/2 |1(5 - 8) + 4(8 - 2) + 7(2 - 5)| Area = 1/2 |1(-3) + 4(6) + 7(-3)| Area = 1/2 |-3 + 24 - 21| Area = 1/2 |0| Area = 0

Since the area is 0, the points P, Q, and R are collinear.

Centroid of a Triangle

The centroid of a triangle is the point of intersection of its medians. A median is a line segment joining a vertex to the midpoint of the opposite side. The centroid is also the center of mass of the triangle.

If the vertices of a triangle are A(x1, y1), B(x2, y2), and C(x3, y3), the coordinates of its centroid G are given by:

G(x, y) = ( (x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3 )

Example: Find the centroid of the triangle with vertices (1, 4), (3, -2), and (-5, 6).

x1 = 1, y1 = 4 x2 = 3, y2 = -2 x3 = -5, y3 = 6

x = (1 + 3 + (-5)) / 3 = (4 - 5) / 3 = -1 / 3

y = (4 + (-2) + 6) / 3 = (4 - 2 + 6) / 3 = 8 / 3

The centroid is (-1/3, 8/3).

Orthocenter of a Triangle

The orthocenter of a triangle is the point of intersection of its altitudes. An altitude is a line segment from a vertex perpendicular to the opposite side.

Finding the orthocenter involves calculating the equations of two altitudes and finding their point of intersection.

Let the vertices be A(x1, y1), B(x2, y2), C(x3, y3).

1. Calculate the slope of side BC (m_BC).
2. The slope of the altitude from A to BC (m_alt_A) will be the negative reciprocal of m_BC (m_alt_A = -1 / m_BC).
3. Write the equation of the altitude from A using the point-slope form: y - y1 = m_alt_A * (x - x1).
4. Repeat steps 1-3 for another vertex, say B, to find the equation of the altitude from B to AC.
5. Solve the system of linear equations formed by the two altitudes to find the coordinates of the orthocenter.

Special Case: In a right-angled triangle, the orthocenter is the vertex where the right angle is formed. In an equilateral triangle, the orthocenter coincides with the centroid and circumcenter.

Circumcenter of a Triangle

The circumcenter is the center of the circle that passes through all three vertices of the triangle (the circumcircle). It is the point of intersection of the perpendicular bisectors of the sides of the triangle.

To find the circumcenter:

1. Find the midpoint of one side (e.g., AB).
2. Calculate the slope of that side (m_AB).
3. The slope of the perpendicular bisector of AB (m_perp_AB) is the negative reciprocal of m_AB.
4. Write the equation of the perpendicular bisector using the midpoint and m_perp_AB.
5. Repeat steps 1-4 for another side (e.g., BC).
6. Solve the system of linear equations formed by the two perpendicular bisectors to find the coordinates of the circumcenter.

Special Case: In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse. In an equilateral triangle, it coincides with the centroid and orthocenter.

Incenter of a Triangle

The incenter is the center of the inscribed circle (incircle) of the triangle. It is the point of intersection of the angle bisectors of the triangle. The incenter is equidistant from the sides of the triangle.

If the vertices of a triangle are A(x1, y1), B(x2, y2), and C(x3, y3), and the lengths of the sides opposite to these vertices are a, b, and c respectively (i.e., a = BC, b = AC, c = AB), then the coordinates of the incenter I are:

I(x, y) = ( (a*x1 + b*x2 + c*x3) / (a + b + c), (a*y1 + b*y2 + c*y3) / (a + b + c) )

Note that the side lengths a, b, c must be calculated using the distance formula first.

Euler's Line

For any non-equilateral triangle, the orthocenter (H), centroid (G), and circumcenter (O) are collinear. This line is called Euler's line. The centroid G lies between H and O, and divides the segment HO in the ratio 2:1 (HG : GO = 2 : 1).

This means that the coordinates of the centroid G can also be found using the section formula if the coordinates of the orthocenter H and circumcenter O are known:

G = (1*H + 2*O) / (1 + 2) = (H + 2*O) / 3

Introduction to Straight Lines

A straight line in a Cartesian plane is a set of points satisfying a linear equation. The general form of the equation of a straight line is Ax + By + C = 0, where A, B, and C are constants, and A and B are not both zero.

Slope of a Line

The slope of a line is a measure of its steepness. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two distinct points on the line. It is usually denoted by 'm'.

If a line makes an angle θ with the positive direction of the x-axis, then its slope is given by:

m = tan(θ)

If two points (x1, y1) and (x2, y2) lie on the line, the slope is:

m = (y2 - y1) / (x2 - x1)

Interpretation of Slope:

• If m > 0, the line rises from left to right (acute angle with the positive x-axis).
• If m < 0, the line falls from left to right (obtuse angle with the positive x-axis).
• If m = 0, the line is horizontal (parallel to the x-axis). The angle θ is 0 degrees.
• If m is undefined, the line is vertical (parallel to the y-axis). The angle θ is 90 degrees, and tan(90°) is undefined.

Parallel Lines: Two non-vertical lines with slopes m1 and m2 are parallel if and only if m1 = m2.

Perpendicular Lines: Two non-vertical lines with slopes m1 and m2 are perpendicular if and only if m1 * m2 = -1 (or one slope is the negative reciprocal of the other).

Equations of Straight Lines in Various Forms

There are several ways to represent the equation of a straight line, depending on the given information.

1. Slope-Intercept Form

If the slope of the line is 'm' and its y-intercept (the y-coordinate where the line crosses the y-axis) is 'c', the equation is:

y = mx + c

The y-intercept 'c' is the y-coordinate of the point (0, c) where the line intersects the y-axis.

2. Point-Slope Form

If a line has slope 'm' and passes through a point (x1, y1), its equation is:

y - y1 = m(x - x1)

3. Two-Point Form

If a line passes through two distinct points (x1, y1) and (x2, y2), its equation can be found using the slope (m = (y2 - y1) / (x2 - x1)) and the point-slope form:

y - y1 = ((y2 - y1) / (x2 - x1)) * (x - x1)

This can be rearranged as:

(y - y1) / (x - x1) = (y2 - y1) / (x2 - x1)

4. Intercept Form

If a line makes x-intercept 'a' and y-intercept 'b' (meaning it crosses the x-axis at (a, 0) and the y-axis at (0, b)), its equation is:

x/a + y/b = 1

Note: 'a' and 'b' must be non-zero.

5. Normal Form

If the perpendicular distance from the origin to the line is 'p' and the angle this perpendicular makes with the positive x-axis is 'α', the equation is:

x cos(α) + y sin(α) = p

This form is particularly useful in certain coordinate geometry problems.

6. General Form (Ax + By + C = 0)

This is the most general form. To convert it to other forms:

• To slope-intercept form (y = mx + c): Rearrange to get y = (-A/B)x - (C/B). So, m = -A/B and c = -C/B. (Assuming B is not 0).
• To intercept form (x/a + y/b = 1): Rearrange to -Ax - By = C. If C is not 0, divide by -C: (Ax/(-C)) + (By/(-C)) = 1. This gives x/(-C/A) + y/(-C/B) = 1. So, a = -C/A and b = -C/B. (Assuming A, B, C are non-zero).

Angle Between Two Lines

If two lines have slopes m1 and m2, the tangent of the angle (θ) between them is given by:

tan(θ) = |(m1 - m2) / (1 + m1*m2)|

This formula gives the tangent of the acute angle between the lines. If 1 + m1*m2 = 0, then m1*m2 = -1, which means the lines are perpendicular, and tan(θ) is undefined (the angle is 90°).

If one of the lines is vertical (slope undefined), say m1 is undefined, and the other line has slope m2, the angle φ it makes with the vertical line is given by cot(φ) = |m2|.

Distance of a Point from a Line

The perpendicular distance 'd' of a point P(x0, y0) from the line Ax + By + C = 0 is given by:

d = |Ax0 + By0 + C| / √(A² + B²)

Example: Find the distance of the point (2, 3) from the line 3x + 4y - 10 = 0.

Here, A = 3, B = 4, C = -10, x0 = 2, y0 = 3.

d = |3(2) + 4(3) - 10| / √(3² + 4²) d = |6 + 12 - 10| / √(9 + 16) d = |8| / √(25) d = 8 / 5 units.

Distance Between Two Parallel Lines

The distance 'd' between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by:

d = |C1 - C2| / √(A² + B²)

Note that the coefficients of x and y (A and B) must be the same for both lines. If they are not, you need to adjust the equations by multiplying them by appropriate constants before applying the formula.

Shifting of Origin

Sometimes, it is convenient to shift the origin to a new point (h, k) without changing the direction of the axes. If the original coordinates of a point are (x, y) and the new coordinates after shifting the origin to (h, k) are (x', y'), the relationship is:

x = x' + h => x' = x - h

y = y' + k => y' = y - k

To shift the origin to a new point (h, k), substitute x = x' + h and y = y' + k into the equation of the curve or line.

Example: Shift the origin to the point (2, 3) and express the equation y - 3 = 2(x - 2) in terms of the new coordinates.

Here, h = 2, k = 3.

The new coordinates are x' = x - 2 and y' = y - 3.

The given equation is already in the form y' = 2x'.

The equation of the line in the new coordinate system is y' = 2x'.

Rotation of Axes

Rotation of axes is used to eliminate the xy term in the equation of a conic section or to simplify equations. If the axes are rotated by an angle θ counterclockwise, the new coordinates (x', y') are related to the old coordinates (x, y) by:

x = x' cos(θ) - y' sin(θ)

y = x' sin(θ) + y' cos(θ)

Conversely:

x' = x cos(θ) + y sin(θ)

y' = -x cos(θ) + y sin(θ)

To eliminate the xy term in an equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0, the angle of rotation θ is given by:

cot(2θ) = (A - C) / B

This is a more advanced topic typically covered in conic sections, but the basic transformation formulas are important.

Summary of Key Concepts

The Cartesian rectangular coordinate system provides a bridge between algebra and geometry. Understanding its components—axes, origin, quadrants, and coordinates—is crucial. Key formulas derived from this system include:

• Distance Formula: d = √((x2 - x1)² + (y2 - y1)²)
• Section Formula (Internal): x = (m*x2 + n*x1) / (m + n), y = (m*y2 + n*y1) / (m + n)
• Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2)
• Area of Triangle: 1/2 |x1(y2 - y3) + ...|
• Centroid: G = ((x1+x2+x3)/3, (y1+y2+y3)/3)
• Slope: m = tan(θ) = (y2 - y1) / (x2 - x1)
• Line Equations: y = mx + c (slope-intercept), y - y1 = m(x - x1) (point-slope), x/a + y/b = 1 (intercept)
• Distance from Point to Line: d = |Ax0 + By0 + C| / √(A² + B²)

Mastering these concepts and formulas will equip you to solve a wide range of problems in coordinate geometry, which are frequently tested in competitive exams like JEE Main.