NCERT Solutions for Class 10 Maths: Coordinate Geometry Exercise 7.1

Understanding the Distance Formula: From Pythagoras to Coordinates

The core of Exercise 7.1 is the Distance Formula. But where does it come from? It's a clever application of the Pythagorean theorem. Imagine you have two points on a graph: P with coordinates (x₁, y₁) and Q with coordinates (x₂, y₂). If you draw a straight line between them, you can form a right-angled triangle with this line as the hypotenuse.

The length of the horizontal side of this triangle is the difference in the x-coordinates, which is |x₂ - x₁|.
The length of the vertical side is the difference in the y-coordinates, which is |y₂ - y₁|.

According to the Pythagorean theorem (a² + b² = c²), we have:
(Horizontal side)² + (Vertical side)² = (Hypotenuse)²
(x₂ - x₁)² + (y₂ - y₁)² = (Distance PQ)²

To find the actual distance PQ, we just take the square root of both sides. This gives us the famous Distance Formula:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

This single formula is the key to solving every problem in this exercise. It allows us to measure the length of any straight line segment on a coordinate plane.

Worked Examples: Applying the Distance Formula

• Example 1: Find the distance between the points A(2, 3) and B(4, 1). Step 1: Identify the coordinates. Let A(2, 3) be (x₁, y₁) and B(4, 1) be (x₂, y₂). So, x₁ = 2, y₁ = 3, x₂ = 4, and y₂ = 1. Step 2: Write down the distance formula. d = √[(x₂ - x₁)² + (y₂ - y₁)²] Step 3: Substitute the values into the formula. d = √[(4 - 2)² + (1 - 3)²] Step 4: Simplify the expression inside the square root. d = √[(2)² + (-2)²] d = √[4 + 4] d = √8 Step 5: Simplify the radical. √8 = √(4 × 2) = 2√2 Final Answer: The distance between points A and B is 2√2 units.
• Example 2: Find the point on the x-axis which is equidistant from A(2, -5) and B(-2, 9). Step 1: Define the point on the x-axis. Any point on the x-axis has its y-coordinate as 0. Let the point be P(x, 0). Step 2: Set up the condition. The problem states that the point P is equidistant from A and B. This means PA = PB. It's easier to work with squares to avoid radicals, so we can write PA² = PB². Step 3: Use the distance formula to find PA² and PB². PA² = (x₂ - x₁)² + (y₂ - y₁)² = (x - 2)² + (0 - (-5))² = (x - 2)² + (5)² PA² = x² - 4x + 4 + 25 = x² - 4x + 29 PB² = (x - (-2))² + (0 - 9)² = (x + 2)² + (-9)² PB² = x² + 4x + 4 + 81 = x² + 4x + 85 Step 4: Equate PA² and PB² and solve for x. x² - 4x + 29 = x² + 4x + 85 Subtract x² from both sides: -4x + 29 = 4x + 85 -4x - 4x = 85 - 29 -8x = 56 x = -7 Final Answer: The required point on the x-axis is (-7, 0).

Exam Tips: Avoid These Common Mistakes!

The distance formula is simple, but small mistakes can lead to wrong answers. Watch out for these common errors:

• Sign Errors with Negative Coordinates: This is the most common mistake! When you subtract a negative coordinate, it becomes addition. For example, if x₁ = -5 and x₂ = 2, then (x₂ - x₁) is (2 - (-5)) = 2 + 5 = 7. A quick calculation might lead you to 2-5=-3, which is incorrect. Double-check your signs.
• Forgetting to Square: The formula is (x₂ - x₁)² + (y₂ - y₁)², not (x₂ - x₁) + (y₂ - y₁). Always square the differences before adding them.
• Forgetting the Final Square Root: The expression inside the radical, (x₂ - x₁)² + (y₂ - y₁)², is the square of the distance. You must take the square root of this value to find the final distance 'd'.
• Mixing up Coordinates: Don't mix x and y values. The formula subtracts x's from x's and y's from y's. A mix-up like (x₂ - y₁)² will give a completely wrong result. Stick to the pattern.

Practice Questions with Solutions

• Q: Find the distance between the points P(-5, 7) and Q(-1, 3). A: Step 1: Identify the coordinates. Let P(-5, 7) be (x₁, y₁) and Q(-1, 3) be (x₂, y₂). x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3. Step 2: Substitute the values into the distance formula d = √[(x₂ - x₁)² + (y₂ - y₁)²]. d = √[(-1 - (-5))² + (3 - 7)²] Step 3: Simplify the expression. d = √[(-1 + 5)² + (-4)²] d = √[(4)² + (-4)²] d = √[16 + 16] = √32 Step 4: Simplify the radical. √32 = √(16 × 2) = 4√2. Final answer: The distance is 4√2 units.
• Q: Determine if the points A(1, 5), B(2, 3), and C(-2, -11) are collinear. A: Step 1: To check for collinearity, we find the lengths of the three segments AB, BC, and AC and see if the sum of two equals the third. Step 2: Calculate distance AB. AB = √[(2-1)² + (3-5)²] = √[1² + (-2)²] = √(1+4) = √5. Step 3: Calculate distance BC. BC = √[(-2-2)² + (-11-3)²] = √[(-4)² + (-14)²] = √(16+196) = √212 = 2√53. Step 4: Calculate distance AC. AC = √[(-2-1)² + (-11-5)²] = √[(-3)² + (-16)²] = √(9+256) = √265. Step 5: Check if AB + BC = AC or any other combination. √5 + 2√53 ≠ √265. Since the sum of the lengths of any two segments is not equal to the length of the third segment, the points are not collinear. Final answer: The points are not collinear.
• Q: Check whether (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle. A: Step 1: Let the points be A(5, -2), B(6, 4), and C(7, -2). An isosceles triangle has at least two sides of equal length. We need to calculate the lengths of AB, BC, and AC. Step 2: Calculate AB. AB = √[(6-5)² + (4 - (-2))²] = √[1² + 6²] = √(1+36) = √37. Step 3: Calculate BC. BC = √[(7-6)² + (-2 - 4)²] = √[1² + (-6)²] = √(1+36) = √37. Step 4: Calculate AC. AC = √[(7-5)² + (-2 - (-2))²] = √[2² + 0²] = √4 = 2. Step 5: Compare the lengths. We see that AB = BC = √37. Since two sides are equal, the triangle is isosceles. Final answer: Yes, the points are the vertices of an isosceles triangle because two of its sides have equal length (√37 units).
• Q: Find the value(s) of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units. A: Step 1: We are given the distance d = 10. The points are P(2, -3) and Q(10, y). Step 2: Use the distance formula: d² = (x₂ - x₁)² + (y₂ - y₁)². 10² = (10 - 2)² + (y - (-3))² Step 3: Simplify and solve for y. 100 = (8)² + (y + 3)² 100 = 64 + (y + 3)² 100 - 64 = (y + 3)² 36 = (y + 3)² Step 4: Take the square root of both sides. Remember to consider both positive and negative roots. √36 = y + 3 ±6 = y + 3 Step 5: Find the two possible values for y. Case 1: 6 = y + 3 => y = 6 - 3 = 3. Case 2: -6 = y + 3 => y = -6 - 3 = -9. Final answer: The possible values for y are 3 and -9.

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