Find all points Q such that
(i) d(P,Q)= 5 when P= (3,-3),
(ii) the x-coordinate of Q = 6
I am wondering what the function d is.
Q would be the infinite set of points (x,y) such that
(x-3)^2 + (y+3)^2 = 25
ii) your second part makes no sense to me.
You wanted Q to be a set of points, thus Q=6 is meaningless
Did you mean, find x when y = 6 ???
To find all points Q that satisfy the given conditions, we need to consider each condition separately.
(i) d(P,Q) = 5 when P = (3, -3):
The distance between two points P(x1, y1) and Q(x2, y2) in a two-dimensional coordinate plane can be calculated using the distance formula:
d(P,Q) = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Using this formula, we can substitute the values P = (3, -3) and d(P,Q) = 5 into the formula and solve for Q.
(sqrt((x2 - 3)^2 + (y2 - (-3))^2)) = 5
Squaring both sides of the equation eliminates the square root:
(x2 - 3)^2 + (y2 + 3)^2 = 25
This is the equation of a circle centered at P = (3, -3) with a radius of 5. To find all points Q on this circle, we can determine the coordinates of the center and the radius.
Center = (3, -3)
Radius = 5
Using these values, we can graph the circle and find all points Q that are 5 units away from P = (3, -3).
(ii) The x-coordinate of Q = 6:
From the condition, we know that the x-coordinate of Q is 6. This means that Q must have the form (6, y), where y can be any real number.
Therefore, all the points Q that satisfy both conditions (i) and (ii) simultaneously are the points on the circle centered at P = (3, -3) with a radius of 5 and with an x-coordinate of 6.
To visualize these points, you can plot the circle centered at (3,-3) with a radius of 5 and mark the point (6, y) on the circle, where y can be any real number.