A point (x,y) is at a distance of 5 units from the origin. How many such points lie in the thirdquadrant

Sure, let's break this down:

1. First, we need to understand what the question is asking. It's asking for points that are 5 units away from the origin (0,0) and lie in the third quadrant of a graph.

2. The third quadrant is where both x and y are negative.

3. The distance from the origin to any point (x,y) in a plane is given by the formula √(x² + y²). Since the distance is given as 5 units, we can set up the equation as follows: √(x² + y²) = 5.

4. Squaring both sides to get rid of the square root gives us x² + y² = 25.

5. Now, we need to find the points that satisfy this equation and lie in the third quadrant.

6. In the third quadrant, both x and y are negative. So, we are looking for pairs of (x, y) that satisfy the equation x² + y² = 25 and where both x and y are negative.

7. There are infinite points that satisfy this condition. For example, (-3,-4) and (-4,-3) are two such points. If you plot these points and all other points that satisfy the condition on a graph, they will form a circle with a radius of 5 units centered at the origin, and the points in the third quadrant will be part of this circle.

So, there are infinite points that are 5 units away from the origin and lie in the third quadrant.