Find the sum of all values of $a$ such that the point $(a,7)$ is $3\sqrt{5}$ from the point $(2,1)$.
+9466 From the Pythagorean theorem...
[ distance between (a, 7) and (2, 1) ]2 = [a - 2]2 + [7 - 1]2
And they tell us that the distance equals 3√5 , so
[ 3√5 ]2 = [a - 2]2 + [7 - 1]2
45 = [a - 2]2 + 36
9 = [a - 2]2
±√9 = a - 2
± 3 = a - 2
± 3 + 2 = a
a = 3 + 2 = 5 or a = -3 + 2 = -1
and
5 + -1 = 4
+9466 From the Pythagorean theorem...
[ distance between (a, 7) and (2, 1) ]2 = [a - 2]2 + [7 - 1]2
And they tell us that the distance equals 3√5 , so
[ 3√5 ]2 = [a - 2]2 + [7 - 1]2
45 = [a - 2]2 + 36
9 = [a - 2]2
±√9 = a - 2
± 3 = a - 2
± 3 + 2 = a
a = 3 + 2 = 5 or a = -3 + 2 = -1
and
5 + -1 = 4
