find all points having an x-coordinate of 4 whose distance from the point (1,5) is 5
you know that all points at a distance 5 will form a circle with center at (1,5) and radius 5. So,
(x-1)^2 + (y-5)^2 = 25
So, plug in x=4 and you have
3^2 + (y-5)^2 = 25
(y-5)^2 = 16
y-5 = ±4
y = 9 or 1
So, the points are (4,9) and (4,1)
Use the distance formula.
which is?
It should be on your formula sheet if not then look it up.
Thanks Steve!
To find all points with an x-coordinate of 4 that are 5 units away from the point (1,5), we can use the distance formula. The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Where (x1, y1) represents the coordinates of the given point (1,5), (x2, y2) represents the coordinates of the point we are trying to find, and d is the distance we want to be equal to 5.
Let's apply this formula using the given values: x1 = 1, y1 = 5, x2 = 4, and d = 5.
5 = sqrt((4 - 1)^2 + (y2 - 5)^2)
Simplifying further:
25 = (4 - 1)^2 + (y2 - 5)^2
25 = 9 + (y2 - 5)^2
Subtracting 9 from both sides:
16 = (y2 - 5)^2
Taking the square root of both sides:
±4 = y2 - 5
Now, solve for y2 by adding 5 to both sides:
y2 = 5 ± 4
This gives us two possible points: (4, 9) and (4, 1).
Therefore, the two points with an x-coordinate of 4 and a distance of 5 from the point (1,5) are (4, 9) and (4, 1).