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

2 Answers
Dec 31, 2017

Answer:

#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#

Dec 31, 2017

Answer:

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]