How do you find a possible value for a if the points (-2,a), (6,1)has a distance of d=4sqrt5?

2 Answers
Jim G.
Dec 31, 2017

a=-3" or "a=5

Explanation:

"using the "color(blue)"distance formula"

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

"let "(x_1,y_1)=(6,1)" and "(x_2,y_2)=(-2,a)

d=sqrt((-2-6)^2+(a-1)^2)=4sqrt5

rArrsqrt(64+(a-1)^2)=4sqrt5

color(blue)"square both sides"

rArr64+(a-1)^2=80

"subtract 64 from both sides"

(a-1)^2=16

color(blue)"take the square root of both sides"

rArra-1=+-sqrt16larrcolor(blue)"note plus or minus"

rArra=1+-4

rArra=1-4=-3" or "a=1+4=5

Binayaka C.
Dec 31, 2017

Possible values of a are -3 and 5.

Explanation:

Distance between two points (x_1,y_1) and (x_2,y_2) is

D= sqrt((x_1-x_2)^2+(y_1-y_2)^2)) Therefore,

Distance between two points (-2,a) and (6,1) is

D= sqrt((-2-6)^2+(a-1)^2)= 4sqrt5 , Squaring both

sides we get , 64+ a^2-2a+1=80 or

a^2-2a+1+64-80=0 or

a^2-2a-15=0 or a^2-5a+3a-15=0 or

a(a-5)+3(a-5)=0 :. (a+3)(a-5)=0

:. a =-3, a = 5

Possible values of a are -3 and 5 [Ans]

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