Class 9 Math Chapter 9 Exercise 9.1 Notes: Calculus and Analytic Geometry Distance Formula | New 2025-2026 Syllabus Solved PDF Notes

What are the Key Concepts and Rules for this Exercise?

Analytical geometry—often referred to as coordinate geometry—serves as the critical bridge between pure algebra and geometric analysis, laying the foundational framework for advanced calculus and analytic geometry in higher secondary and university mathematics. Introduced by René Descartes, this branch of mathematics utilizes a rectangular coordinate system (Cartesian plane) to represent geometric figures algebraically.

In the New 2025-2026 Single National Curriculum (SNC) textbook for Class 9 Mathematics, Chapter 9: Introduction to Coordinate Geometry focuses on establishing quantitative measurements between points on a plane. The key theoretical principles governing Exercise 9.1 include:

• The Cartesian Coordinate Plane: A two-dimensional plane defined by two mutually perpendicular real number lines intersecting at an origin $O(0,0)$. The horizontal line is the $x$-axis, and the vertical line is the $y$-axis. Any point $P$ in the plane is defined uniquely by an ordered pair $(x, y)$.
• Derivation of the Distance Formula: Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be any two distinct points in the plane. By constructing a right-angled triangle $PRQ$ where $R$ has coordinates $(x_2, y_1)$, the horizontal distance is $|x_2 - x_1|$ and the vertical distance is $|y_2 - y_1|$. Applying the Pythagorean theorem to $\triangle PRQ$: $$d = |PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Since squaring any real number yields a non-negative result, $(x_2 - x_1)^2 = (x_1 - x_2)^2$, meaning the order of points does not affect the final distance.
• Collinear Points: Three or more points $A$, $B$, and $C$ are said to be collinear if they lie on the exact same straight line. Geometrically, if $B$ lies between $A$ and $C$, then $|AB| + |BC| = |AC|$. If $|AB| + |BC| \neq |AC|$, the points form a non-collinear set and define the vertices of a triangle.
• Geometric Shape Verification:
  • Equilateral Triangle: All three sides are equal ($|AB| = |BC| = |CA|$).
  • Isosceles Triangle: Any two sides are equal in length (e.g., $|AB| = |AC| \neq |BC|$).
  • Scalene Triangle: All three sides have distinct lengths ($|AB| \neq |BC| \neq |CA|$).
  • Right-Angled Triangle: The lengths satisfy the Pythagorean relation ($|AB|^2 + |BC|^2 = |AC|^2$).

What is the Formula Reference Table for this Chapter?

Below is the essential formula reference table required for solving problems in Chapter 9 (Analytic Geometry) under the 2025-2026 Board Syllabus:

Concept / Property	Mathematical Formula / Condition	Application Scope
Distance Formula	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	Distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$
Distance from Origin	$d = \sqrt{x^2 + y^2}$	Distance from $O(0,0)$ to point $P(x,y)$
Points on Axis ($x$-axis & $y$-axis)	$d = \sqrt{a^2 + b^2}$	Distance between $P(a, 0)$ on $x$-axis and $Q(0, b)$ on $y$-axis
Collinearity Test	$|AB| + |BC| = |AC|$	Checking if three points $A, B, C$ lie on a straight line
Right Triangle Test	$|AB|^2 + |BC|^2 = |AC|^2$	Verifying right-angled triangle via Pythagorean converse
Midpoint Formula	$M(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$	Finding the exact middle point of segment $PQ$

How to Solve All Exercise Questions Step-by-Step? (Complete Solved Solutions)

Below are the full mathematical derivations and solved problems for Exercise 9.1 (Class 9 Mathematics, New 2025-2026 Textbook Standard).

Question 1: Find the distance between the following pairs of points.

(a) $A(2, 3)$ and $B(6, 6)$

Solution:

Let $x_1 = 2, y_1 = 3$ and $x_2 = 6, y_2 = 6$.

Using the Distance Formula:

$$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substitute the given values into the formula:

$$|AB| = \sqrt{(6 - 2)^2 + (6 - 3)^2}$$ $$|AB| = \sqrt{(4)^2 + (3)^2}$$ $$|AB| = \sqrt{16 + 9}$$ $$|AB| = \sqrt{25} = 5\text{ units}$$

Final Answer: The distance $|AB|$ is $5$ units.

(b) $A(-4, 1)$ and $B(3, -2)$

Solution:

Let $x_1 = -4, y_1 = 1$ and $x_2 = 3, y_2 = -2$.

Applying the distance formula:

$$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$|AB| = \sqrt{(3 - (-4))^2 + (-2 - 1)^2}$$ $$|AB| = \sqrt{(3 + 4)^2 + (-3)^2}$$ $$|AB| = \sqrt{(7)^2 + 9}$$ $$|AB| = \sqrt{49 + 9}$$ $$|AB| = \sqrt{58}\text{ units}$$

Final Answer: The distance $|AB|$ is $\sqrt{58}$ units.

(c) $A(-8, 1)$ and $B(6, 1)$

Solution:

Let $x_1 = -8, y_1 = 1$ and $x_2 = 6, y_2 = 1$.

Applying the distance formula:

$$|AB| = \sqrt{(6 - (-8))^2 + (1 - 1)^2}$$ $$|AB| = \sqrt{(6 + 8)^2 + (0)^2}$$ $$|AB| = \sqrt{(14)^2 + 0}$$ $$|AB| = \sqrt{196} = 14\text{ units}$$

Alternative Analysis: Since both points share the same $y$-coordinate ($y_1 = y_2 = 1$), the segment is strictly horizontal. Thus, $|AB| = |x_2 - x_1| = |6 - (-8)| = |14| = 14$ units.

Final Answer: The distance $|AB|$ is $14$ units.

(d) $A(-4, \sqrt{2})$ and $B(4, -3\sqrt{2})$

Solution:

Let $x_1 = -4, y_1 = \sqrt{2}$ and $x_2 = 4, y_2 = -3\sqrt{2}$.

Applying the distance formula:

$$|AB| = \sqrt{(4 - (-4))^2 + (-3\sqrt{2} - \sqrt{2})^2}$$ $$|AB| = \sqrt{(4 + 4)^2 + (-4\sqrt{2})^2}$$ $$|AB| = \sqrt{(8)^2 + ((-4)^2 \cdot (\sqrt{2})^2)}$$ $$|AB| = \sqrt{64 + (16 \cdot 2)}$$ $$|AB| = \sqrt{64 + 32}$$ $$|AB| = \sqrt{96}$$

Simplifying the radical expression:

$$|AB| = \sqrt{16 \times 6} = 4\sqrt{6}\text{ units}$$

Final Answer: The distance $|AB|$ is $4\sqrt{6}$ units.

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Question 2: Let $P$ be a point on the $x$-axis with $x$-coordinate $a$ and $Q$ be a point on the $y$-axis with $y$-coordinate $b$. Find the distance between $P$ and $Q$.

General Algebraic Formulation:

Since $P$ lies on the $x$-axis, its coordinates are $P(a, 0)$.
Since $Q$ lies on the $y$-axis, its coordinates are $Q(0, b)$.

Using the distance formula:

$$|PQ| = \sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{(-a)^2 + (b)^2} = \sqrt{a^2 + b^2}$$

(i) $a = 6, b = 8$

Solution:

$$|PQ| = \sqrt{(6)^2 + (8)^2}$$ $$|PQ| = \sqrt{36 + 64}$$ $$|PQ| = \sqrt{100} = 10\text{ units}$$

Final Answer: $|PQ| = 10$ units.

(ii) $a = -9, b = 12$

Solution:

$$|PQ| = \sqrt{(-9)^2 + (12)^2}$$ $$|PQ| = \sqrt{81 + 144}$$ $$|PQ| = \sqrt{225} = 15\text{ units}$$

Final Answer: $|PQ| = 15$ units.

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Question 3: Show whether the points $A(1, 2)$, $B(4, 6)$, and $C(7, 10)$ are collinear using the distance formula.

Solution:

To test collinearity, we compute the lengths of all three line segments: $|AB|$, $|BC|$, and $|AC|$.

Step 1: Calculate $|AB|$

$$|AB| = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 2: Calculate $|BC|$

$$|BC| = \sqrt{(7 - 4)^2 + (10 - 6)^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 3: Calculate $|AC|$

$$|AC| = \sqrt{(7 - 1)^2 + (10 - 2)^2} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Step 4: Check collinearity condition $|AB| + |BC| = |AC|$

$$|AB| + |BC| = 5 + 5 = 10$$ $$|AC| = 10$$

Since $|AB| + |BC| = |AC|$, point $B$ lies on the line segment joining $A$ and $C$.

Conclusion: The points $A(1, 2)$, $B(4, 6)$, and $C(7, 10)$ are collinear.

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Question 4: Verify whether the vertices $P(-2, 3)$, $Q(2, 2)$, and $R(1, -2)$ form an isosceles triangle or a right-angled triangle.

Solution:

We need to determine the lengths of sides $|PQ|$, $|QR|$, and $|PR|$.

Step 1: Distance $|PQ|$

$$|PQ| = \sqrt{(2 - (-2))^2 + (2 - 3)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$$

Step 2: Distance $|QR|$

$$|QR| = \sqrt{(1 - 2)^2 + (-2 - 2)^2} = \sqrt{(-1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$$

Step 3: Distance $|PR|$

$$|PR| = \sqrt{(1 - (-2))^2 + (-2 - 3)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$$

Step 4: Analysis of geometric properties

• Isosceles Property: Since $|PQ| = \sqrt{17}$ and $|QR| = \sqrt{17}$, two sides are equal ($|PQ| = |QR|$). Thus, $\triangle PQR$ is an isosceles triangle.
• Right-Angled Property (Pythagorean Theorem): $$|PQ|^2 + |QR|^2 = (\sqrt{17})^2 + (\sqrt{17})^2 = 17 + 17 = 34$$ $$|PR|^2 = (\sqrt{34})^2 = 34$$ Since $|PQ|^2 + |QR|^2 = |PR|^2$, the converse of Pythagorean theorem holds true, meaning the angle at $Q$ is $90^\circ$.

Conclusion: The given points $P$, $Q$, and $R$ form an isosceles right-angled triangle.

Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)

Frequently Asked Questions: What are Common Student Errors in this Exercise?

Q1: Why do students lose marks when dealing with negative coordinates in the distance formula?
Ans: The most frequent mistake in board exams (FBISE & Punjab Boards) occurs when substituting negative coordinates into $(x_2 - x_1)^2$. For example, if $x_1 = -4$ and $x_2 = 3$, students often write $(3 - 4)^2$ instead of $(3 - (-4))^2 = (3 + 4)^2 = 7^2 = 49$. Always use parentheses around negative values to avoid sign errors.

Q2: Can the distance between two points in analytic geometry ever be negative?
Ans: No. Distance is a scalar magnitude representing spatial separation and is defined via a square root $\sqrt{\text{expression}} \ge 0$. Even if the coordinate differences are negative, squaring them yields positive quantities. Therefore, geometric distance is strictly non-negative ($d \ge 0$).

Q3: How does Chapter 9 Coordinate Geometry connect to advanced Calculus and Analytic Geometry in FSC / University level?
Ans: Coordinate geometry in Class 9 introduces the Cartesian plane, distance evaluation, and midpoint computation. In higher-level calculus and analytic geometry (such as F.Sc Math Part 2 Chapter 4 and university calculus), these exact principles are expanded to define vector magnitudes, conic sections (parabolas, ellipses, circles), derivatives as slopes of tangents, and definite integrals as area bounded by coordinate curves.