How do you find a possible value for a if the points (a,0), (3,1) has a distance of #d=sqrt2#?

1 Answer
May 14, 2017

Answer:

#a=2# or #a=4#

Explanation:

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

Hence distance between #(a,0)# and #3,1# is

#sqrt((a-3)^2+(0-1)^2)=sqrt(a^2-6a+9+1)=sqrt(a^2-6a+10)#

as this is equal to #sqrt2#, we have

#a^2-6a+10=2# or #a^2-6a+8=0#

or #a^2-4a-2a+8=0#

i.e. #a(a-4)-2(a-4)=0#

or #(a-2)(a-4)=0#

i.e. #a=2# or #a=4#

graph{((x-2)^2+y^2-0.01)((x-4)^2+y^2-0.01)((x-3)^2+(y-1)^2-0.01)=0 [-3.11, 6.89, -1.92, 3.08]}