Find two points, one on the horizontal axis and one on the vertical axis, such that the distance between these two points equals 15.
NEED HELP PLEASE!!
How about (0,12) and (9,0) ?
There are other possibilities also.
yes, these two points does equal to 15, but how did you find these points?
I thought of creating a 3:4:5 (ratio of sides) right triangle, with the hypotenuse equal to 15. The 3:4:5 ratio stays the same.
I could have also chosen
(0, 15/sqrt2) and (15/sqrt2, 0)
or
(-3,0) and (0,-4)
and many others.
To find two points, one on the horizontal axis and one on the vertical axis, such that the distance between these two points equals 15, we need to understand the concept of the distance formula.
The distance formula is derived from the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse's length is equal to the sum of the squares of the lengths of the other two sides. In our case, the hypotenuse is the distance between the two points.
The distance formula is given by d = √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two points.
Since we want one point on the horizontal axis, we can assume its y-coordinate to be zero. Let's call its x-coordinate x1. Similarly, since we want one point on the vertical axis, we can assume its x-coordinate to be zero. Let's call its y-coordinate y2.
Substituting these values into the distance formula, we get:
15 = √((0 - x1)^2 + (y2 - 0)^2)
15 = √(x1^2 + y2^2)
Since one of the coordinates is zero, we can simplify the equation further:
15 = √(x1^2 + y2^2)
15 = √(x1^2 + 0^2)
15 = √x1^2
15 = |x1|
From this equation, we can see that the absolute value of x1 must be 15. So x1 can be either 15 or -15.
Therefore, we have two points that satisfy the given conditions: (15, 0) and (-15, 0). The first point lies on the positive side of the horizontal axis, and the second point lies on the negative side of the horizontal axis. These two points have a distance of 15 units between them.