Question

If the distance between the points A(2,-2) and B(-1, x) is equal to 5, then
the value of x is:
(b)-2
(c)1
(d)-1​

Answer 1

Answer:

GivenAB=5

\implies AB^{2} = 5^{2}⟹AB2=52

\implies (x_{2} - x_{1})^{2} + (y_{2}-y_{1})^{2}=25⟹(x2−x1)2+(y2−y1)2=25

\blue{ ( By \: Distance \:formula . ) }(ByDistanceformula.)

\implies ( -1-2)^{2} + [x-(-2)]^{2} = 25⟹(−1−2)2+[x−(−2)]2=25

\implies (-3)^{2} + (x+2)^{2} = 25⟹(−3)2+(x+2)2=25

\implies 9 + (x+2)^{2} = 25⟹9+(x+2)2=25

\implies (x+2)^{2} = 25 - 9⟹(x+2)2=25−9

\implies (x+2)^{2} = 16⟹(x+2)2=16

\implies (x+2)= \pm \sqrt{16 }⟹(x+2)=±16

\implies x + 2 = \pm 4⟹x+2=±4

\implies x = -2 \pm 4⟹x=−2±4

\implies x = -2 + 4 \:or \: x = -2 - 4⟹x=−2+4orx=−2−4

\implies x = 2 \:or \: x = -6⟹x=2orx=−6