Question

Determinati valorile reale ale numarului a, stiind ca distanta dintre punctele A(3, 2) si B( a, 8) este egala cu 10.

Answer 1

AB=√(Xb-Xa)²+(Yb-Ya)²
AB²=(Xb-Xa)²+(Yb-Ya)²
100=(a-3)²+(8-2)²
a²-6a+9+36=100
a²-6a-55=0
Δ=(-6)²-4×(-55)=256
a₁=(6+√256)/2=(6+16)/2=22/2=11
a₂=(6-√256)/2=(6-16)/2=-10/2=-5

Answer 2

apliocam formula distantei intre 2 puncte

(3-a)²+(2-8)²=10²
(3-a)²=100-36=64

deci
 3-a=8
3-8=a
-5=a
 a=-5

sau
3-a=-8
3+8=a
11=a
a=11
 deci a∈{-5;11}

sau, altfel
a²-6a+9=64
a²-6a-55=0
a1=11
a2=-5