Chapter – 7

Coordinate Geometry

In this post we have given the detailed notes of class 10 Maths Chapter 7 (Coordinate Geometry) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 7: Coordinate Geometry

What is Coordinate Geometry?

Coordinate geometry (also called analytical geometry) is the branch of mathematics that connects algebra and geometry using a coordinate system. The position of a point in a plane is described using an ordered pair of numbers called the coordinates of that point.

Coordinate geometry lets us find the distance between two points, divide a line segment in a given ratio, find the mid-point of a line segment, and calculate the area of a triangle formed by three points — all using algebraic formulas instead of actual drawing and measurement.

The Cartesian Plane

• XOX′ is the horizontal line, called the x-axis.
• YOY′ is the vertical line, called the y-axis.
• The point O, where the x-axis and y-axis intersect, is called the origin. Its coordinates are (0, 0).
• The x-axis and y-axis divide the plane into four regions called quadrants.

The Four Quadrants and Their Signs

• Quadrant I (XOY): both coordinates positive → (+, +)
• Quadrant II (X′OY): x negative, y positive → (−, +)
• Quadrant III (X′OY′): both coordinates negative → (−, −)
• Quadrant IV (XOY′): x positive, y negative → (+, −)

Points on the Axes

• Any point on the positive x-axis has coordinates (x, 0), where x > 0.
• Any point on the negative x-axis has coordinates (x, 0), where x < 0.
• Any point on the positive y-axis has coordinates (0, y), where y > 0.
• Any point on the negative y-axis has coordinates (0, y), where y < 0.
• The origin has coordinates (0, 0).

Note: Every point on the x-axis has its y-coordinate equal to zero, and every point on the y-axis has its x-coordinate equal to zero.

Abscissa and Ordinate

• Abscissa (x-coordinate) — the distance of a point from the y-axis.
• Ordinate (y-coordinate) — the distance of a point from the x-axis.
• The coordinates of a point are written as an ordered pair (x, y), where x is the abscissa and y is the ordinate.

Distance Formula

The distance between two points is the length of the line segment joining them. If P(x₁, y₁) and Q(x₂, y₂) are any two points in a plane, the distance between them is given by:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

This is called the Distance Formula. Since distance is always non-negative, we always take the positive square root.

Distance of a point from the origin: The distance of a point P(x, y) from the origin O(0, 0) is:

OP = √(x² + y²)

Applications of the Distance Formula

• To check whether three given points are collinear.
• To check whether a given set of points forms a particular type of triangle (isosceles, equilateral, right-angled, or scalene).
• To check whether a given set of points forms a special quadrilateral (square, rectangle, rhombus, or parallelogram).

Section Formula

The section formula is used to find the coordinates of a point that divides the line segment joining two given points in a given ratio.

If a point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂, then the coordinates of P are:

P(x, y) = [ (m₁x₂ + m₂x₁)/(m₁ + m₂) , (m₁y₂ + m₂y₁)/(m₁ + m₂) ]

This is derived using similar triangles (△PAQ ~ △BPC), formed by drawing perpendiculars from A, P and B onto the x-axis.

Mid-point Formula

The mid-point of a line segment divides it in the ratio 1 : 1. Substituting m₁ = m₂ = 1 in the section formula, the coordinates of the mid-point of the line segment joining A(x₁, y₁) and B(x₂, y₂) are:

Mid-point = [ (x₁ + x₂)/2 , (y₁ + y₂)/2 ]

Worked Example — Section Formula

Q. Find the coordinates of the point which divides the line segment joining the points (4, −3) and (8, 5) internally in the ratio 3 : 1.

Solution: Let P(x, y) be the required point. Using the section formula with m₁ = 3, m₂ = 1:

x = {3(8) + 1(4)}/(3 + 1) = (24 + 4)/4 = 7

y = {3(5) + 1(−3)}/(3 + 1) = (15 − 3)/4 = 3

Hence, the required point is (7, 3).

Area of a Triangle

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle, its area is given by the numerical value of the expression:

Area of △ABC = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

Since area can never be negative, we always take the numerical (absolute) value of the result.

Worked Example — Area of a Triangle

Q. Find the area of the triangle whose vertices are (1, −1), (−4, 6) and (−3, −5).

Solution: Using the area formula with A(1, −1), B(−4, 6), C(−3, −5):

Area = ½ [1(6 − (−5)) + (−4)(−5 − (−1)) + (−3)(−1 − 6)]

= ½ [1(11) + (−4)(−4) + (−3)(−7)]

= ½ (11 + 16 + 21) = ½ (48) = 24 square units

Worked Example — Collinear Points

Q. Find the value of k, if the points A(2, 3), B(4, k) and C(6, −3) are collinear.

Solution: Since the three points are collinear, the area of the triangle formed by them is 0.

½ [2(k + 3) + 4(−3 − 3) + 6(3 − k)] = 0

½ (2k + 6 − 24 + 18 − 6k) = 0

½ (−4k + 0) = 0 ⟹ k = 0

Hence, the required value of k is 0.

Worked Example — Area of a Quadrilateral

Q. If A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5) are the vertices of a quadrilateral ABCD, find its area.

Solution: Join B to D. This divides quadrilateral ABCD into two triangles, ABD and BCD.

Area of △ABD = ½ [−5(−5 − 5) + (−4)(5 − 7) + 4(7 + 5)] = ½ (50 + 8 + 48) = 53 square units

Area of △BCD = ½ [−4(−6 − 5) + (−1)(5 + 5) + 4(−5 + 6)] = ½ (44 − 10 + 4) = 19 square units

Area of quadrilateral ABCD = Area of △ABD + Area of △BCD = 53 + 19 = 72 square units

Worked Example — Medians of a Triangle (Centroid)

Q. A(4, 2), B(6, 5) and C(1, 4) are the vertices of a triangle ABC. (i) Find the coordinates of the point D where the median through A meets BC. (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1. (iii) Find the coordinates of points Q and R on medians BE and CF such that BQ : QE = 2 : 1 and CR : RF = 2 : 1. What do you observe?

Solution: D is the mid-point of BC, so D = ((6+1)/2, (5+4)/2) = (7/2, 9/2)

Using the section formula, P on AD with AP : PD = 2 : 1: P = {(2×7/2 + 1×4)/3, (2×9/2 + 1×2)/3} = (11/3, 11/3)

E is the mid-point of AC, so E = ((4+1)/2, (2+4)/2) = (5/2, 3). Q on BE with BQ : QE = 2 : 1: Q = {(2×5/2 + 1×6)/3, (2×3 + 1×5)/3} = (11/3, 11/3)

F is the mid-point of AB, so F = ((4+6)/2, (2+5)/2) = (5, 7/2). R on CF with CR : RF = 2 : 1: R = {(2×5 + 1×1)/3, (2×7/2 + 1×4)/3} = (11/3, 11/3)

We observe that P, Q and R are the same point, (11/3, 11/3). This point is called the centroid of the triangle — the point where all three medians meet, and it always divides each median in the ratio 2 : 1 from the vertex.

Key Points to Remember

• Distance Formula: PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
• Distance of a point from the origin: OP = √(x² + y²)
• Section Formula (internal division in ratio m₁ : m₂): [ (m₁x₂ + m₂x₁)/(m₁+m₂) , (m₁y₂ + m₂y₁)/(m₁+m₂) ]
• Mid-point Formula: [ (x₁+x₂)/2 , (y₁+y₂)/2 ]
• Area of a Triangle: ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
• If the area of a triangle formed by three points is 0, the points are collinear.
• The centroid of a triangle divides each median in the ratio 2 : 1 from the vertex.
• Any point on the x-axis has coordinates (x, 0); any point on the y-axis has coordinates (0, y).
• Area is always taken as a positive (numerical/absolute) value.

We hope that class 10 Maths Chapter 7 (Coordinate Geometry) notes in English helped you. If you have any query about class 10 Maths Chapter 7 (Coordinate Geometry) notes in English or about any other notes of class 10 Maths in English, so you can comment below. We will reach you as soon as possible…