Find the sum of all values of a such that the point (a,7) is 3 sqrt 5 from the point (2,1).
up to the left, and up to the right
d^2 = (2-a)^2 + (1-7)^2
d^2 = 45 = 4 - 4 a + a^2 + 36
5 = a^2 - 4 a
a^2 - 4 a - 5 = 0
(a-5)(a+1) = 0
a = 5 or a = -1
Or, you can think of it as looking for the intersections of a circle and a line.
The circle: (x-2)^2 + (y-1)^2 = 45
The line: y=7
They intersect where
(x-2)^2 + (7-1)^2 = 45
now the math proceeds as above
To solve this problem, we need to find the values of "a" such that the distance between the point (a, 7) and the point (2, 1) is equal to 3√5.
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Using this formula, we can set up the equation:
3√5 = √((2 - a)^2 + (1 - 7)^2)
Simplifying this equation, we get:
45 = (2 - a)^2 + 36
Combining like terms, we have:
45 = (a - 2)^2 + 36
Now, let's solve this equation to find the values of "a".
45 - 36 = (a - 2)^2
9 = (a - 2)^2
Taking the square root of both sides, we have:
±√9 = ±(a - 2)
Simplifying further, we get:
±3 = a - 2
Now, we solve for "a".
a - 2 = 3: a = 3 + 2 = 5
a - 2 = -3: a = -3 + 2 = -1
Therefore, the values of "a" that satisfy the given condition are 5 and -1.
To find the sum of these two values, we calculate:
5 + (-1) = 4
So, the sum of all values of "a" is 4.