Use of Coordinate Geometry Class 9: Concept, Formulas & Examples | Video Solutions

Use of Coordinate Geometry Class 9: Cartesian System, Distance Formula & Midpoint Formula Explained

Coordinate Geometry Class 9 (Chapter 1, “Orienting Yourself: The Use of Coordinates”) teaches you how to locate any point on a plane using an ordered pair (x, y), understand the four quadrants and their signs, and use the Distance Formula and Midpoint Formula to solve Coordinate Geometry problems using coordinates instead of just diagrams.

Watch the Use of Coordinate Geometry Class 9 Full Video Explanation on Youtube:

Reading through Distance Formula derivations like the one in Coordinate Geometry Class 9 can feel abstract on a page. Watching it explained step-by-step Video, exactly as shown in this Akash Institute class, makes the logic & Concept much Easier — especially the right-triangle-based derivation and the room-floor-plan case study.

What Is the Cartesian Coordinate System in Coordinate Geometry Class 9?

The Cartesian coordinate system is the method used in Coordinate Geometry Class 9 to describe the exact location or position of any point on a flat, two-dimensional surface. To locate any point in Coordinate Geometry, you first need a fixed reference point — this reference point is called the origin. From the origin, two perpendicular lines are drawn: a horizontal line called the x-axis and a vertical line called the y-axis.

Once these two axes are drawn, any point’s position can be described by measuring its distance along the x-axis and along the y-axis. These two distances, written together, are called the coordinates of the point — written as (a, b). The x-coordinate is also known as the abscissa, and the y-coordinate is also known as the ordinate. This entire system, where two perpendicular lines meeting at a point are used to fix positions, is called the Cartesian Coordinate System — the foundation of Coordinate Geometry Class 9.

What Are Quadrants and How Do the Signs Work in Coordinate Geometry Class 9?

When the x-axis and y-axis intersect, they form four 90° angles, adding up to a full 360°. Each one-fourth part of this Cartesian plane is called a quadrant. There are four quadrants in total in Coordinate Geometry, and every quadrant has a fixed sign pattern for the coordinates of any point lying inside it.

This quadrant sign rule from Coordinate Geometry Class 9 is frequently tested in a slightly tricky way. For example: if a point is given as (−a, b) and it lies in Quadrant II, students often assume “a” is negative just because of the minus sign written before it. But since Quadrant II already requires the x-coordinate to be negative, and the coordinate given is “−a”, this means “a” itself must be positive for “−a” to come out negative. Similarly, “b” is already positive as given. So in this Coordinate Geometry Class 9 example, both a and b turn out to be positive values.

How Are Coordinates Used to Solve Real-Life Problems in Coordinate Geometry Class 9?

Q1. On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (-7, 0) to (13, 0) on the x-axis and from (0, -15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using, answer the given questions.

Place Reigan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7).

Coordinate Geometry Class 9 becomes much easier to understand once you see it applied to a real floor-plan style problem, exactly like the one solved in the reference video using a room layout marked on a graph sheet, with the x-axis running from −7 to 13 and the y-axis running from −15 to 12.

(i) Where will the fourth foot of the table be?

Solution:  A rectangular study table has three of its feet given at (8, 9), (11, 9), and (11, 7). Since a rectangle’s opposite sides must be equal, the fourth foot is found at (8, 7).

(ii) Is this a good spot for the table?

Solution:  This is a good spot for the table because it doesn’t block the wardrobe or any door, and it sits right against the wall.

(iii) What is the width of the table? The length?

Solution: The width of the table is the gap between (11,9) and (11,7), which is 2 units. The length is the gap between (8,9) and (11,9), which is 3 units.

(iv) Can you make out the height of the table?

Solution: The height of the table cannot be calculated from this figure, because it is drawn in 2D — there is no third dimension shown to measure height in this Cartesian plane diagram.

(v) If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe?

Solution: The bathroom door’s hinge point is at (0, 1.5), and it opens up to the point (0, 4). Checking the gap (2.5 units) against the wardrobe’s position confirms that the door does not hit the wardrobe when opened.

(vi) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.

Solution: A shower area named SHWR, with corners S(−6, 6), H(−3, 6), W(−2, 9) and R(−6, 9), has one pair of opposite sides equal — this identifies its shape as a trapezium.

How Do You Find the Distance Between Two Points in Coordinate Geometry Class 9?

The Distance Formula is one of the most important tools in Coordinate Geometry Class 9. It gives you the straight-line distance between any two points P(x₁, y₁) and Q(x₂, y₂) on the Cartesian plane. The Distance Formula is derived by drawing perpendiculars from both points to form a right-angled triangle, where the horizontal leg equals (x₂ − x₁) and the vertical leg equals (y₂ − y₁).

Applying the Pythagoras theorem to this right triangle gives the Distance Formula used throughout Coordinate Geometry Class 9.

Q2. Case Study: Deriving the Distance Formula Using Right Triangle PQR

The figure shows two points P(x₁, y₁) and Q(x₂, y₂) plotted in the first quadrant of a rectangular coordinate system. Perpendiculars are drawn from P and Q to the axes, and a horizontal line from P meets the vertical line through Q at point R, forming a right-angled triangle PQR. Using this figure, derive the Distance Formula step by step.

Solution:

• Step 1: Identify the coordinates of point R.
  Point R is formed where the horizontal line from P meets the vertical line from Q. Since R lies on the same horizontal level as P, its y-coordinate is y₁. Since R lies on the same vertical line as Q, its x-coordinate is x₂. So, the coordinates of R are (x₂, y₁).
• Step 2: Find the length of PR.
  PR is the horizontal distance between P(x₁, y₁) and R(x₂, y₁). Since both points lie on the same horizontal line, PR = x₂ − x₁.
• Step 3: Find the length of QR.
  QR is the vertical distance between Q(x₂, y₂) and R(x₂, y₁). Since both points lie on the same vertical line, QR = y₂ − y₁.
• Step 4: Identify the right angle in triangle PQR.
  PR is a horizontal segment and QR is a vertical segment, so PR is perpendicular to QR. This means angle R = 90°, and PQ is the hypotenuse of the right-angled triangle PQR.
• Step 5: Apply the Pythagoras theorem.
  In a right-angled triangle, (hypotenuse)² = (base)² + (perpendicular)². Applying this to triangle PQR:
  PQ² = QR² + PR²
• Step 6: Substitute the values of QR and PR.
  PQ² = (y₂ − y₁)² + (x₂ − x₁)²
• Step 7: Take the square root of both sides to find PQ.
  PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
• Step 8: Write the final Distance Formula.
  This gives the Distance Formula between any two points P(x₁, y₁) and Q(x₂, y₂):
  PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

This formula can also be written as PQ = √[(x₁ − x₂)² + (y₁ − y₂)²], since squared terms are always positive, regardless of the order of subtraction.

Why this diagram helps: It shows visually why the Distance Formula works — it isn’t just a formula to memorise, but a direct application of the Pythagoras theorem to a right triangle formed by the horizontal gap (x₂ − x₁) and vertical gap (y₂ − y₁) between any two points.

Distance Formula: PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

This Distance Formula can also be written as √[(x₁ − x₂)² + (y₁ − y₂)²], since squared terms are always positive regardless of the order of subtraction.

Solved Example: Distance Formula in Coordinate Geometry Class 9

Q3. Find the distance between (4, 6) and (−2, 9) using the Distance Formula.

Q4. Find the distance between the points –  R(a + b, a − b) and S(a − b, −a − b)

A slightly harder algebraic version of the same Distance Formula idea uses variables instead of numbers: for points (a+b, a−b) and (−a, −b), applying the Distance Formula and simplifying gives a distance of 2√(a² + b²) units.

How to Find the Area of a Triangle Using Coordinate Geometry Class 9?

Q5. Find the area of the triangle formed by O(0, 0), A(4, 1) and B(−2, −5) such that the length of the perpendicular from the origin to AB is 3/√2​ units.

1. When you’re given the base as a distance between two coordinate points and the perpendicular height, you can find the area of a triangle using the standard formula, Area = ½ × base × height.
2. For example, with O(0,0), A(4,1), and B(−2,−5), and a given perpendicular height from the origin to AB of 3/√2 units, the base AB is first calculated using the Distance Formula as 6√2 units.
3. Applying the area formula: Area = ½ × 6√2 × 3/√2 = 9 square units. Area answers in Coordinate Geometry Class 9 are always written in square units, such as cm² or m².

How Do You Prove a Shape Is a Square Using the Distance Formula?

Q6. If the points A(0, 0), B(1, 0), C(1, 1) and D(0, 1) are the vertices of a quadrilateral, then find out which special type of quadrilateral it is?

To prove that four given points form a square in Coordinate Geometry Class 9, you calculate all four side lengths using the Distance Formula and check if they are equal, then calculate both diagonals using the same Distance Formula to check if they’re also equal.

NOTE: If all sides come out equal but the diagonals are not equal, the shape is a rhombus instead of a square. Once confirmed as a square, its area is simply side × side.

How Do You Prove a Right-Angled Triangle Using Coordinate Geometry?

Q7. Prove that (4, 4), (3, 5) and (−1, −1) are the vertices of a right-angled triangle.

A set of three points forms a right-angled triangle if the square of the longest side (hypotenuse) equals the sum of the squares of the other two sides, following the Pythagoras theorem.

For example, with A(4,4), B(3,5), and C(−1,−1):

1. AB works out to √2 units,
2. BC to √52 units, and
3. AC to √50 units, (all calculated using the Distance Formula).

Checking BC² = AC² + AB² gives 52 = 50 + 2 = 52, which confirms the triangle is right-angled, with BC as the hypotenuse.

If, in addition to this, two of the sides are also equal in length (AB = BC) and (AC² = AB²+BC²), the triangle is called a right-angled isosceles triangle.

What Is the Condition for Collinear Points in Coordinate Geometry Class 9?

Three points A, B, and C are called collinear points if they all lie on the exact same straight line. Collinear points don’t form any triangle or shape with area — their enclosed area is always zero. The condition to check collinear points is: the sum of two smaller line segments must equal the third, larger segment — for example, AB + BC = AC.

In a solved example with A(0,4), B(−1, 16/3), and C(2, 4/3): AB works out to 5/3 units, BC to 5 units, and AC to 10/3 units — all found using the Distance Formula. Since AB + AC = 5/3 + 10/3 = 5, which equals BC, the three points satisfy the collinear points condition and are confirmed to be collinear.

What Is the Midpoint Formula in Coordinate Geometry Class 9?

The Midpoint Formula is another key formula in Coordinate Geometry Class 9. The midpoint of a line segment joining two points A(x₁, y₁) and B(x₂, y₂) is the point that divides the segment exactly in half (a 1:1 ratio). Its coordinates are found using the Midpoint Formula by averaging the x-coordinates and averaging the y-coordinates.

Midpoint Formula: ( (x₁+x₂)/2 , (y₁+y₂)/2 )

Solved Example: Using the Midpoint Formula in Coordinate Geometry Class 9

1. Find the midpoint of A(2, −7) and B(9, −2) using the Midpoint Formula: Midpoint = (11/2, −9/2).
2. Now find the distance between this midpoint and the point C(5, −6) using the Distance Formula.
3. Substituting and simplifying gives a final distance of 5/2 units.

A more advanced application of the Midpoint Formula in Coordinate Geometry Class 9 is finding the original vertices of a triangle when only the midpoints of its sides are given. For instance, if the midpoints of the sides of triangle ABC are D(5,1), E(6,5), and F(0,3), setting up and solving simultaneous equations using the Midpoint Formula for the x-coordinates and y-coordinates separately gives the triangle’s actual vertices as A(−1,−1), B(11,3), and C(1,7).

Practice Problems: Circles and Case-Study Questions in Coordinate Geometry Class 9

• An engineering “nut” diagram case study uses front view, top view, and side view to test Coordinate Geometry Class 9 skills. In the front view, point B is at (2, 6), and point H’s distance from the y-axis is 7 units.
• In the top view of the same diagram, point A is at (2,3) and point B at (3,1); the distance AB, found using the Distance Formula, works out to √5 units.
• Also in the top view, points J(3,5) and C(6,1) have a midpoint of (4.5, 3), calculated using the Midpoint Formula.
• In a circle-based Coordinate Geometry Class 9 problem, points A(1,−8), B(−4,7), and C(−7,−4) are checked against the origin as centre: OA, OB, and OC (all via the Distance Formula) work out to √65 units, confirming all three points lie on a circle of radius √65 units.
• Checking point D(−5, 6) against this same circle using the Distance Formula: OD works out to √61 units, which is less than the radius √65 units, so point D lies inside the circle.

Frequently Asked Questions on Coordinate Geometry Class 9

What is the Cartesian Coordinate System in Coordinate Geometry Class 9?

It is a system where two perpendicular lines — the x-axis and y-axis — intersect at a fixed reference point called the origin, allowing any point’s position in Coordinate Geometry Class 9 to be described as an ordered pair of coordinates (x, y), also called the abscissa and ordinate.

What are the signs of coordinates in each quadrant of Coordinate Geometry Class 9?

In Coordinate Geometry Class 9, Quadrant I has both coordinates positive (+,+), Quadrant II has (−,+), Quadrant III has (−,−), and Quadrant IV has (+,−), based on the four 90° divisions created where the axes intersect.

What is the Distance Formula in Coordinate Geometry Class 9?

The Distance Formula, PQ = √[(x₂−x₁)² + (y₂−y₁)²], is derived using the Pythagoras theorem on a right triangle formed by the horizontal and vertical differences between two points, and it gives the straight-line distance between them in Coordinate Geometry Class 9.

How do you check if three points are collinear using the Distance Formula?

Calculate the distances between all pairs of the three points using the Distance Formula. If the sum of the two smaller distances equals the largest distance (for example, AB + BC = AC), the three points lie on the same line and are collinear points.

What is the Midpoint Formula used for in Coordinate Geometry Class 9?

The Midpoint Formula, ( (x₁+x₂)/2, (y₁+y₂)/2 ), finds the exact centre point of a line segment joining two coordinates, and is also used to work backward and find a triangle’s vertices when only the midpoints of its sides are known — a common application in Coordinate Geometry Class 9.

What is the difference between abscissa and ordinate in Coordinate Geometry Class 9?

The abscissa is the x-coordinate of a point, showing its distance from the y-axis, while the ordinate is the y-coordinate, showing its distance from the x-axis — together they form the coordinates (abscissa, ordinate) of a point in Coordinate Geometry Class 9.