In the xy-coordinate plane, the distznce between point B(15,7) and point A(x,15) is 17. What is one possible value of x?
To find one possible value of x, we can use the distance formula:
The distance formula is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Given that point B is (15,7) and the distance between point B and point A(x,15) is 17, we can plug in the values into the distance formula.
The x-coordinate of point A is x, and the y-coordinate of point A is 15.
So, substituting the values into the distance formula, we have:
17 = √((x - 15)^2 + (15 - 7)^2)
Simplifying, we get:
289 = (x - 15)^2 + 64
Expanding the equation further:
289 = x^2 - 30x + 225 + 64
Combining like terms:
289 = x^2 - 30x + 289
Subtracting 289 from both sides of the equation:
0 = x^2 - 30x
Now, we can factor out x:
0 = x(x - 30)
Setting each factor equal to zero:
x = 0 or x - 30 = 0
From this, we get two possible values for x:
x = 0 or x = 30
So, one possible value of x is 0 and another possible value is 30.