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4.1 Q-3

Question Statement

Determine which of the following points are at a distance of 15 units from the origin:

  1. (176,7)(\sqrt{176}, 7)
  2. (10,βˆ’10)(10, -10)
  3. (1,15)(1, 15)
  4. (152,152)\left(\frac{15}{2}, \frac{15}{2}\right)

Background and Explanation

To solve this problem, we calculate the distance of each point from the origin (0,0)(0, 0) using the distance formula:

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Here, (x1,y1)(x_1, y_1) represents the origin (0,0)(0, 0), and (x2,y2)(x_2, y_2) is the given point.
We will substitute the coordinates of each point and determine if the distance is exactly 15.


Solution

1. Point: (176,7)(\sqrt{176}, 7)

Using the distance formula:

d=(176βˆ’0)2+(7βˆ’0)2d = \sqrt{\left(\sqrt{176} - 0\right)^2 + (7 - 0)^2} d=176+49=225=15d = \sqrt{176 + 49} = \sqrt{225} = 15

Conclusion: The point (176,7)(\sqrt{176}, 7) is at a distance of 15 units from the origin.


2. Point: (10,βˆ’10)(10, -10)

Using the distance formula:

d=(10βˆ’0)2+(βˆ’10βˆ’0)2d = \sqrt{(10 - 0)^2 + (-10 - 0)^2} d=100+100=200=102d = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}

Conclusion: The point (10,βˆ’10)(10, -10) is not at a distance of 15 units from the origin.


3. Point: (1,15)(1, 15)

Using the distance formula:

d=(1βˆ’0)2+(15βˆ’0)2d = \sqrt{(1 - 0)^2 + (15 - 0)^2} d=1+225=226d = \sqrt{1 + 225} = \sqrt{226}

Conclusion: The point (1,15)(1, 15) is not at a distance of 15 units from the origin.


4. Point: (152,152)\left(\frac{15}{2}, \frac{15}{2}\right)

Using the distance formula:

d=(152βˆ’0)2+(152βˆ’0)2d = \sqrt{\left(\frac{15}{2} - 0\right)^2 + \left(\frac{15}{2} - 0\right)^2} d=2254+2254=4504=112.5d = \sqrt{\frac{225}{4} + \frac{225}{4}} = \sqrt{\frac{450}{4}} = \sqrt{112.5}

Conclusion: The point (152,152)\left(\frac{15}{2}, \frac{15}{2}\right) is not at a distance of 15 units from the origin.


Key Formulas or Methods Used

  1. Distance Formula:
d=(x2βˆ’x1)2+(y2βˆ’y1)2 d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  1. Substitution: The origin (0,0)(0, 0) simplifies calculations.

  2. Square Roots: Simplify carefully to compare the result to 15.


Summary of Steps

  1. Identify coordinates of each point and the origin (0,0)(0, 0).
  2. Apply the distance formula to calculate the distance for each point.
  3. Simplify the result and compare it to 15.
  4. Conclude whether each point satisfies the condition of being at a distance of 15 units.

Final Results

  • (176,7)(\sqrt{176}, 7): At a distance of 15 units.
  • (10,βˆ’10)(10, -10): Not at a distance of 15 units.
  • (1,15)(1, 15): Not at a distance of 15 units.
  • (152,152)\left(\frac{15}{2}, \frac{15}{2}\right): Not at a distance of 15 units.