Point (x, y) is at distance of 5 units from the origin. How many such points lie in the third quadrant?

The distance of a point (x, y) from the origin is given by the formula:

d = √(x² + y²)

Here, the distance is given as 5 units. Therefore, we have:

√(x² + y²) = 5

Square both sides to eliminate the square root:

x² + y² = 25

This equation represents a circle with radius 5 centered at the origin.

Step 1: Analyze the third quadrant
In the third quadrant, both x and y are negative. Thus, any point (x, y) in the third quadrant must satisfy:
– x < 0
– y < 0
– x² + y² = 25

Step 2: Check if there are infinitely many solutions
For any point on the circle x² + y² = 25, there are infinitely many points that satisfy this equation because it is a continuous curve. Specifically, in the third quadrant, there are infinitely many points where both x and y are negative, as long as they satisfy the circle equation.

For example:
– If x = -3, then y² = 25 – (-3)² = 25 – 9 = 16, so y = -4 (since y < 0 in the third quadrant).
– If x = -4, then y² = 25 – (-4)² = 25 – 16 = 9, so y = -3 (since y < 0 in the third quadrant).

This process can be repeated for infinitely many values of x in the range -5 ≤ x < 0, each corresponding to a unique y-value in the third quadrant.

Step 3: Conclusion
There are infinitely many points in the third quadrant that lie on the circle x² + y² = 25.

The correct answer is:
d) infinitely many
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.

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