Question 6 of 18intermediate🔧 ApplyShort Answer3 marks

Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Correct Answer

Let the point on the x-axis be P(x, 0). Since this point is equidistant from A(2, -5) and B(-2, 9), the distances PA and PB must be equal. We use the condition PA² = PB² to avoid square roots.

Using the distance formula, we set up the equation:

(x−2)2+(0−(−5))2=(x−(−2))2+(0−9)2(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2 (x−2)2+25=(x+2)2+81(x - 2)^2 + 25 = (x + 2)^2 + 81

Expanding and simplifying the terms:

x2−4x+4+25=x2+4x+4+81x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 −4x+29=4x+85-4x + 29 = 4x + 85 −8x=56-8x = 56 x=−7x = -7

The required point on the x-axis is (-7, 0).

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

Explanation

The solution applies the logic from Example 5 in the context, adapting it for the x-axis. Just as a point on the y-axis is of the form (0, y), a point on the x-axis is of the form (x, 0). By equating the squares of the distances from the unknown point to the two given coordinates, we solve algebraically for the x-coordinate.

Solution Steps

  1. Step 1: Let the point on the x-axis be P(x, 0).

  2. Step 2: Equate the squared distances: (x - 2)² + (5)² = (x + 2)² + (-9)².

  3. Step 3: Expand the equation: x² - 4x + 4 + 25 = x² + 4x + 4 + 81.

  4. Step 4: Simplify to find x: -4x + 29 = 4x + 85, which gives x = -7.

  5. Step 5: Conclude the point is (-7, 0).

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