Question 49 of 70intermediate🔧 ApplyShort Answer3 marks

Find a point on the xx-axis, which is equidistant from the points (7, 6) and (3, 4).

Correct Answer

The point on the x-axis which is equidistant from (7, 6) and (3, 4) is (15/2, 0) or (7.5, 0).

Exercise: EXERCISE 9.1 | Q: 4 | (Chapter: 8)
For More Understanding

Explanation

This question tests the application of the distance formula in coordinate geometry. A point on the x-axis has the form (x, 0). Using the distance formula, we set the distances from this point to both given points equal to each other and solve for x. The context mentions the distance formula as an important formula from coordinate geometry.

Solution Steps

  1. Step 1: Let the point on x-axis be P(x, 0), since any point on x-axis has y-coordinate = 0.

  2. Step 2: Using distance formula, distance from P to (7, 6) = √[(x-7)² + (0-6)²] = √[(x-7)² + 36]

  3. Step 3: Distance from P to (3, 4) = √[(x-3)² + (0-4)²] = √[(x-3)² + 16]

  4. Step 4: Since P is equidistant from both points: √[(x-7)² + 36] = √[(x-3)² + 16]

  5. Step 5: Squaring both sides: (x-7)² + 36 = (x-3)² + 16

  6. Step 6: Expanding: x² - 14x + 49 + 36 = x² - 6x + 9 + 16

  7. Step 7: Simplifying: -14x + 85 = -6x + 25, which gives 8x = 60, so x = 15/2

  8. Step 8: Therefore, the required point is (15/2, 0).

PreviousNext