NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3
Question 2

Question. 2

Find the point(s) on the x–axis at a distance \(2\sqrt5\) from \((7,-4)\). How many such points are there?

Answer:

(5, 0) and (9, 0) — two points.

Detailed Answer with Explanation:

Step 1: Recall that any point on the x–axis has the form \((x, 0)\), because its y–coordinate is always zero.

Step 2: Use the distance formula. The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 3: Here, one point is \((x, 0)\) and the other is \((7, -4)\). So the distance becomes: \[ d = \sqrt{(x - 7)^2 + (0 - (-4))^2} \]

Step 4: Simplify the second term: \(0 - (-4) = 4\). So, \[ d = \sqrt{(x - 7)^2 + (4)^2} \]

Step 5: The problem says the distance is \(2\sqrt{5}\). That means: \[ \sqrt{(x - 7)^2 + 16} = 2\sqrt{5} \]

Step 6: Square both sides to remove the square root: \[ (x - 7)^2 + 16 = (2\sqrt{5})^2 \]

Step 7: Calculate the right-hand side: \((2\sqrt{5})^2 = 4 \times 5 = 20\). So, \[ (x - 7)^2 + 16 = 20 \]

Step 8: Subtract 16 from both sides: \[ (x - 7)^2 = 20 - 16 = 4 \]

Step 9: Take square root of both sides: \[ x - 7 = \pm 2 \]

Step 10: Solve for \(x\): - If \(x - 7 = 2\), then \(x = 9\). - If \(x - 7 = -2\), then \(x = 5\).

Step 11: So the points are \((9, 0)\) and \((5, 0)\).

Final Answer: There are two points on the x–axis: (5, 0) and (9, 0).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 7: Coordinate Geometry – Exercise 7.3 | Detailed Answers