Find all the points having an x-coordinate of 9 whose distance from the point (3, -2) is 10.
(9, 6), (9, -10)
(9, 13), (9, -7)
(9, -12), (9, 8)
(9, 2), (9, -4)
So ll the points whose distance from the point (3, -2) is 10 would be the circle
(x-3)^2 + (y+2)^2 = 100
when x = 9
36 + (y+2)^2 = 100
(y+2)^2 = 64
y + 2 = ± 8
y = 6 or y = -10
looks like choice #1
To find all the points with an x-coordinate of 9 that are at a distance of 10 from the point (3, -2), we first need to understand the concept of distance between two points in a coordinate system.
The distance between two points (x1, y1) and (x2, y2) can be found using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, we know that the x-coordinate of the points we are trying to find is 9. Let's substitute the known values into the distance formula and solve for y.
d = 10
x1 = 3
y1 = -2
x2 = 9
y2 = y (unknown)
10 = sqrt((9 - 3)^2 + (y - (-2))^2)
Let's simplify the equation:
10 = sqrt(6^2 + (y + 2)^2)
10 = sqrt(36 + (y + 2)^2)
100 = 36 + (y + 2)^2
(y + 2)^2 = 100 - 36
(y + 2)^2 = 64
y + 2 = ±√64
y + 2 = ±8
Now we can solve for y:
If y + 2 = 8, then y = 8 - 2 = 6.
If y + 2 = -8, then y = -8 - 2 = -10.
So the points with an x-coordinate of 9 that are at a distance of 10 from the point (3, -2) are:
(9, 6) and (9, -10).
Therefore, the correct answer is:
(9, 6), (9, -10)