Distance Formula Problems: Solved Examples Class 10

Hey there, future math whiz! Ever looked at a graph with points and wondered, "How far apart are these, yaar?" Or maybe your teacher threw terms like 'midpoint' and 'section formula' at you, and you felt a bit lost? Don't worry, you're not alone!

Coordinate Geometry is super important, not just for your Class 10 board exams but for understanding the world around you. It’s like giving addresses to points on a map and then figuring out relationships between them. Sounds cool, right? Let's dive in and make it super easy for you!

Before we tackle problems, let's quickly recap the powerful formulas that are your best friends in Coordinate Geometry. These are the backbone of the entire chapter, and knowing them inside out is non-negotiable for both CBSE and ICSE students.

### 1. The Distance Formula
This one helps you find the distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in a coordinate plane. Imagine you're trying to find the shortest path between two cities on a grid map. That's exactly what this does!

### 2. The Section Formula
Accha, what if a point divides a line segment in a certain ratio? Like, say, a milestone is placed on a road dividing the path between two towns. The section formula helps you find the coordinates of that point! If point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$, then:

### 3. The Midpoint Formula
This is a special case of the section formula where the ratio is $1:1$. It simply finds the exact middle point of a line segment. Super handy!

### 4. Area of a Triangle using Coordinates
Suno, if you know the coordinates of the three vertices of a triangle, you don't need to measure its base and height! This formula gives you the area directly. If the vertices are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, then:

You might be thinking, "Is this just for exams?" Bilkul nahi! Coordinate Geometry is everywhere. Think about how Google Maps works, it uses coordinates to pinpoint locations and calculate distances. Ever played a video game? The movement of characters and objects on screen is all thanks to coordinates.

Architects use it to design buildings, engineers use it to plan roads and bridges, and even pilots use it for navigation. In fact, fields like data science heavily rely on understanding spatial relationships, and guess what? India's AI market is projected to reach $17 billion by 2027 (NASSCOM), a field where coordinate geometry foundations are super important for things like image recognition and robotics. So, you're not just learning math; you're building skills for the future!

Alright, enough talk! Let's put these formulas to work with some typical Class 10 problems. These examples are similar to what you'd find in your NCERT, RD Sharma, Selina Concise, or RS Aggarwal textbooks and are designed to cover common board exam question patterns. Coordinate Geometry has a weightage of 6 marks in CBSE Class 10, so mastering these types of problems is crucial.

### Example 1: Finding Distance and Collinearity
Question: Determine if the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are collinear.

Solution:
To check for collinearity, we can use the distance formula. If three points $A$, $B$, and $C$ are collinear, then the sum of the distances of any two segments must be equal to the length of the third segment (e.g., $AB + BC = AC$).

Let $A = (1, 5)$, $B = (2, 3)$, and $C = (-2, -11)$.

Step 1: Calculate the distance AB.

Step 2: Calculate the distance BC.

Step 3: Calculate the distance AC.

Step 4: Check for collinearity.
For $A, B, C$ to be collinear, $AB + BC = AC$ (or any permutation).
Is $\sqrt{5} + \sqrt{212} = \sqrt{265}$?
Since $\sqrt{5} \approx 2.23$, $\sqrt{212} \approx 14.56$, and $\sqrt{265} \approx 16.27$.
$2.23 + 14.56 = 16.79$, which is not equal to $16.27$.

Conclusion: Since $AB + BC
eq AC$, the points$(1, 5)$,$(2, 3)$, and$(-2, -11)$ are not collinear.

### Example 2: Using the Section Formula
Question: Find the coordinates of the point which divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.

Solution:
Let the given points be $A(x_1, y_1) = (4, -3)$ and $B(x_2, y_2) = (8, 5)$.
The ratio is $m_1 : m_2 = 3 : 1$.
Let the point dividing the segment be $P(x, y)$.

Step 1: Apply the section formula for x-coordinate.

Step 2: Apply the section formula for y-coordinate.

Conclusion: The coordinates of the point which divides the line segment in the ratio $3:1$ internally are $(7, 3)$.

### Example 3: Area of a Triangle
Question: Find the area of the triangle whose vertices are $(1, -1)$, $(-4, 6)$, and $(-3, -5)$.

Solution:
Let the vertices be $A(x_1, y_1) = (1, -1)$, $B(x_2, y_2) = (-4, 6)$, and $C(x_3, y_3) = (-3, -5)$.

Step 1: Use the area of triangle formula.

Step 2: Substitute the coordinates into the formula.

Conclusion: The area of the triangle with the given vertices is $24$ square units.

Let's quickly sum up what we've covered:

* Distance Formula: Calculates the length of a line segment between two points.
* Section Formula: Finds the coordinates of a point dividing a line segment in a given ratio.
* Midpoint Formula: A special case of the section formula for a 1:1 ratio.
* Area of Triangle Formula: Determines the area of a triangle given its vertices.
* Practice is Paramount: Consistent practice with solved examples is the only way to master these concepts.
* Real-World Relevance: Coordinate geometry is fundamental in many modern applications, from navigation to engineering.

Keep practicing, keep exploring, and you'll ace Coordinate Geometry!