How do you find the possible values for a if the points (a,6), (-6,2) has a distance of #4sqrt10#?
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See a solution process below:
The formula for calculating the distance between two points is:
#d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)#
Substituting the values for the distance and from the points in the problem gives:
#4sqrt(10) = sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2)#
We can now solve for #a#:
First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:
#(4sqrt(10))^2 = (sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2))^2#
#16 * 10 = (color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2#
#160 = (color(red)(-6) - color(blue)(a))^2 + (-4)^2#
#160 = (color(red)(-6) - color(blue)(a))^2 + 16#
#160 = (-6)^2 + 6a + 6a + a^2 + 16#
#160 = 36 + 12a + a^2 + 16#
#160 = a^2 + 12a + 52#
#160 - color(red)(160) = a^2 + 12a + 52 - color(red)(160)#
#0 = a^2 + 12a -108#
We can factor this as:
#0 = (a + 18)(a - 6)
Now, equate each term on the right side of the equation to #0# and solve for #a#:
Solution 1)
#a + 18 = 0#
#a + 18 - color(red)(18) = 0 - color(red)(18)#
#a + 0 = -18#
#a = -18#
Solution 3)
#a - 6 = 0#
#a - 6 + color(red)(6) = 0 + color(red)(6)#
#a - 0 = 6#
#a = 6#
The solution is, a could be #-18# or #6#
Distance between two points, #(x_1,y_1)# and #(x_2,y_2)#
= #sqrt((x_1-x_2)^2+(y_1-y_2)^2#
In this case
#(a,6)# and #(-6,2)#
#4sqrt(10)=sqrt([a-(-6)]^2+(6-2)^2)#
#(4sqrt(10))^2={sqrt((a+6)^2+(4)^2)}^2#
#16*10=(a+6)^2+16#
#160=a^2+6x+6x+36+16#
#a^2+12x-108=0#
Carry out factorisation
#a^2+12x-108=0#
#(a+18)(a-6)=0#
Divide RHS #(0)# with #(a+18)# and #(a-6)#
You will have
#(a+18)=0# and #(a-6)=0# which give you
#a=-18# or #a=6# respectively
