Point A is located at (-3, -5). If point B is 11 units away and is located in quadrant 2, what are the coordinates for point B?
Point B could be any point on the circle given by
(x+3)^2 + (y+5)^2 = 121
if you want a point with integer values it would be (-3,6)
all other points in quad II would have irrational coordinates
e.g. (-2, 2√30 - 5) could be B
To find the coordinates of point B, we need to consider two things: the distance from point A to B and the location of point B in quadrant 2.
Since point A is located at (-3, -5), we know that it is in quadrant 3, as the x-coordinate (-3) is negative and the y-coordinate (-5) is also negative.
Now, let's focus on the distance from point A to B. You mentioned that point B is 11 units away. This means the distance between point A and B is 11 units.
In quadrant 2, the x-coordinate is positive while the y-coordinate is negative. So, to find the coordinates of point B, we need to move 11 units in a positive x-direction and 11 units in a negative y-direction from point A.
Starting from point A (-3, -5), we move 11 units to the right (positive x) to reach the point (8, -5). Now, we move 11 units downwards (negative y) from this point to get to the final point.
Adding the y-coordinate for the downward movement, we get -5 - 11 = -16.
Therefore, the coordinates of point B in quadrant 2 will be (8, -16).
To find the coordinates for point B, you can consider the distance formula.
The distance formula is given by:
d = √( (x2 - x1)^2 + (y2 - y1)^2 )
Let's assume the coordinates of point B as (x, y).
Given that point B is located in quadrant 2, the x-coordinate will be negative and the y-coordinate will be positive.
Given that point B is 11 units away from point A, we can set up the following equation:
11 = √( (x - (-3))^2 + (y - (-5))^2 )
Simplifying the equation, we have:
11 = √( (x + 3)^2 + (y + 5)^2 )
Now, we can square both sides of the equation to eliminate the square root:
11^2 = (x + 3)^2 + (y + 5)^2
Simplifying further:
121 = x^2 + 6x + 9 + y^2 + 10y + 25
Combining like terms:
x^2 + y^2 + 6x + 10y + 155 = 0
Since point B is located in quadrant 2, the x-coordinate will be negative, and the y-coordinate will be positive. This means that the only possible value for x is -4.
Substituting x = -4 into the equation, we have:
(-4)^2 + y^2 + 6(-4) + 10y + 155 = 0
16 + y^2 - 24 + 10y + 155 = 0
y^2 + 10y + 147 = 0
To solve this quadratic equation, we can use the quadratic formula or factorization.
Using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For the equation y^2 + 10y + 147 = 0, the coefficients are:
a = 1, b = 10, c = 147
Substituting these values into the quadratic formula:
y = (-10 ± √(10^2 - 4(1)(147))) / (2(1))
y = (-10 ± √(100 - 588)) / 2
y = (-10 ± √(-488)) / 2
Since the value under the square root is negative, there are no real solutions for y. Therefore, there are no coordinates for point B that satisfy the given conditions.