Question

Q(–1, a) is 4 √5 units away from P(3, –2). Find all possible values of a.

Answer 1

Answer:

a=6,-10

Step-by-step explanation:

$4\sqrt{5}=\sqrt{(3+1)^2+(-2-a)^2} \\squaring\\80=16+(2+a)^2\\80-16=(2+a)^2\\(a+2)^2=64\\taking~square ~root\\\\a+2=\pm 8\\either \\a+2=8,\\a=8-2=6\\or\\a+2=-8\\a=-8-2=-10$