+1800 The distance between the points A and B is √5. If A=(a,−1) and B=(2,2a−7), then find the sum of all possible values of a.
+789 Let's start by plugging the coordinates ofAandBinto the distance formula:
d(A, B) = \sqrt{(2-a)^2 + (-1-2a-7)^2}
We know thatd(A,B)=5, so we can substitute this into the equation to get:
\sqrt{(2-a)^2 + (-1-2a-7)^2} = \sqrt{5}
Squaring both sides of the equation, we get:
(2-a)^2 + (-1-2a-7)^2 = 5
Expanding the parentheses, we get:
4-4a+a^2 + 1+4a+49 = 5
Simplifying, we get:
a^2 - 4a + 50 = 0
Factoring the quadratic, we get:
(a-10)(a-5) = 0
Therefore,a=10ora=5.
Since we are asked for the sum of all possible values ofa, the sum is 10+5=15.
