Question 9 of 18intermediate🔧 ApplyShort Answer3 marks

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

Correct Answer

3x + y = 5

Exercise: EXERCISE 7.1 | Q: 10 | (Page 7)
For More Understanding

Explanation

This question is similar to Example 4 in the textbook. The student needs to apply the distance formula and equate the distances from point P(x,y) to both given points. The key concept is that if a point is equidistant from two points A and B, then AP = BP, which means AP² = BP². After expanding and simplifying, the relation 3x + y = 5 is obtained, which represents the perpendicular bisector of the line segment joining the two given points.

Solution Steps

  1. Step 1: Let P(x, y) be equidistant from A(3, 6) and B(-3, 4). Since AP = BP, we have AP² = BP².

  2. Step 2: Apply distance formula: (x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

  3. Step 3: Expand both sides: x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16

  4. Step 4: Simplify by cancelling x² and y²: -6x - 12y + 45 = 6x - 8y + 25

  5. Step 5: Rearrange: -12x - 4y = -20, which gives 3x + y = 5

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