How do you find the possible values for a if the points (5,8), (a,2) has a distance of #3sqrt5#?

1 Answer
Apr 16, 2017

Answer:

Use the distance formula and solve for #a#:
#{a:a=2,8}#

Explanation:

The distance between the two points #(x_1,y_1)# and #(x_2,y_2)# is given as #d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#.

For this problem, it is given that #(x_1,y_1)=(5,8)#, #(x_2,y_2)=(a,2)#, and #d=3sqrt5#. We need to solve for #a# by plugging these values into the distance formula.
#3sqrt5=sqrt((a-5)^2+(2-8)^2)#.

To solve for #a#, begin by squaring both sides of the equation and simplify a bit.
#(3sqrt5)^color(green)2=sqrt((a-5)^2+(2-8)^2)^color(green)2#
#3^2sqrt5^2=sqrt((a-5)^2+(2-8)^2)^2#
#9*5=(a-5)^2+(2-8)^2#
#45=(a-5)^2+("-"6)^2#
#45=a^2-10a+25+36#
#45=a^2-10a+61#

Now subtract 45 from both sides to get an equation in quadratic form.
#45color(green)(-45)=a^2-10a+61color(green)(-45)#
#0=a^2-10a+16#

Finally, solve for #a# using either the quadratic formula or the factoring method. I'll be factoring.
#0=a^2-2a-8a+16#
#0=a(a-2)-8(a-2)#
#0=(a-8)(a-2)#
#a=8or2#

Therefore the possible values for #a# are defined by the set #{a:a=2,8}#