Q(2, a) is 5 units away from P(–2, –1). Find all possible values of a.

a = 2

Step-by-step explanation:

Here, we want to find all the possible values of a

If Q is 5 units away from P, then the distance between Q and P is 5 units

Mathematically, to find the distance between two points, we use the following formula

D = √( (x2-x1)^2 + (y2-y1)^2

So we have ;

D = √(2-(-2)^2 + (a-(-1)^2

D = √(16) + (a + 1)^2

Since D is 5, we have

5 = √(16) + (a +1)^2

Squaring both sides to remove the square root, we have

25 = 16 + (a + 1)^2

(a + 1)^2 = 25 - 16

(a + 1)^2 = 9

(a + 1)^2 = 3^2

Since power is equal, equate base

a + 1 = 3

a = 3-1

a = 2

throw the computer

step-by-step explanation:

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