Find every point on the horizontal line
y = 5
that is a distance of 5 units away from the point
(6, 1).
To find every point on the horizontal line y = 5 that is a distance of 5 units away from the point (6, 1), we can follow these steps:
Step 1: Write down the equation of the horizontal line: y = 5. This equation represents all points (x, y) on the line where y is always equal to 5.
Step 2: Calculate the x-coordinates of the points that are a distance of 5 units away from the point (6, 1). Since we are looking for points on a horizontal line, the y-coordinates will always be 5. We can use the distance formula to find the x-coordinates.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In our case, x1 = 6, y1 = 1, and y2 = 5. Plugging in these values, we have:
5 = sqrt((x2 - 6)^2 + (5 - 1)^2)
Simplifying, we get:
25 = (x2 - 6)^2 + 16
25 - 16 = (x2 - 6)^2
9 = (x2 - 6)^2
Now, we solve for x2:
Taking the square root of both sides, we get:
x2 - 6 = ±3
Adding 6 to both sides, we have:
x2 = 3 + 6 = 9
or
x2 = -3 + 6 = 3
So, the two x-coordinates of the points that are a distance of 5 units away from (6, 1) on the horizontal line y = 5 are x = 9 and x = 3.
Step 3: Combine the x-coordinates with the y-coordinate of the horizontal line y = 5. Since we are looking for points on this horizontal line, the y-coordinate will always be 5. Therefore, the two points are:
(9, 5) and (3, 5).
These are the two points on the horizontal line y = 5 that are a distance of 5 units away from the point (6, 1).
To find every point on the horizontal line y = 5 that is 5 units away from the point (6, 1), we need to consider the coordinates on the line that are 5 units away horizontally from the given point.
Since the line is horizontal and is defined by y = 5, the y-coordinate of every point on the line is 5. So, we only need to find the x-coordinates of the points that are 5 units away horizontally from (6, 1).
To determine the x-coordinates, we can use the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is given by:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, we have:
distance = 5 units
(x1, y1) = (6, 1)
(x2, y2) = (x, 5) where x represents the x-coordinate of the points we are interested in.
Using this information, we can set up the distance formula equation as follows:
5 = sqrt((x - 6)^2 + (5 - 1)^2)
Squaring both sides of the equation to eliminate the square root gives:
25 = (x - 6)^2 + 16
Expanding the equation further:
25 = x^2 - 12x + 36 + 16
Combining like terms:
25 = x^2 - 12x + 52
Rearranging the equation to bring all terms to one side:
x^2 - 12x + 52 - 25 = 0
x^2 - 12x + 27 = 0
This is a quadratic equation. To solve it, we can either factor it or use the quadratic formula. However, upon inspection, it can be seen that the quadratic cannot be easily factored.
So, let's use the quadratic formula to find the solutions for x:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case:
a = 1, b = -12, c = 27
Plugging these values into the quadratic formula:
x = (-(-12) ± sqrt((-12)^2 - 4(1)(27))) / (2(1))
Simplifying:
x = (12 ± sqrt(144 - 108)) / 2
x = (12 ± sqrt(36)) / 2
x = (12 ± 6) / 2
x1 = (12 + 6) / 2 = 18 / 2 = 9
x2 = (12 - 6) / 2 = 6 / 2 = 3
So, the x-coordinates of the points on the horizontal line y = 5 that are 5 units away from the point (6, 1) are x = 9 and x = 3.
Therefore, the two points on the line are (9, 5) and (3, 5).
To summarize: The points on the horizontal line y = 5 that are a distance of 5 units away from the point (6, 1) are (9, 5) and (3, 5).
well, you know that the vertical distance from (6,1) to y=5 is 4.
So, using a 3,4,5 triangle, the two points are 3 horizontal units away from (6,5), at
(3,5) and (9,5)
If you use your distance formula, you can verify that both of these points are 5 units away from (6,1).