How do you find a possible value for a if the points (a,-6), (-5,2) has a distance of #d=10#?

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 from the points in the problem gives:

#10 = sqrt((color(red)(-5) - color(blue)(a))^2 + (color(red)(2) - color(blue)(-6))^2)#

#10 = sqrt((color(red)(-5) - color(blue)(a))^2 + (color(red)(2) + color(blue)(6))^2)#

#10^2 = (sqrt((color(red)(-5) - color(blue)(a))^2 + (color(red)(2) + color(blue)(6))^2))^2#

#100 = (color(red)(-5) - color(blue)(a))^2 + (color(red)(2) + color(blue)(6))^2#

#100 = (color(red)(-5) - color(blue)(a))^2 + 8^2#

#100 = (color(red)(-5) - color(blue)(a))^2 + 64#

#100 - color(red)(64) = (color(red)(-5) - color(blue)(a))^2 + 64 - color(red)(64)#

#36 = (color(red)(-5) - color(blue)(a))^2 + 0#

#36 = (color(red)(-5) - color(blue)(a))^2#

#sqrt(36) = sqrt((color(red)(-5) - color(blue)(a))^2)#

#6 = +-(-5 - a)#

Solution 1)

#6 = -(-5 - a)#

#6 = 5 + a#

#-color(red)(5) + 6 = -color(red)(5) + 5 + a#

#1 = 0 + a#

#1 = a#

#a = 1#

Solution 2)

#6 = -5 - a#

#color(red)(5) + 6 = color(red)(5) - 5 - a#

#11 = 0 - a#

#11 = -a#

#color(red)(-1) xx 11 = color(red)(-1) xx -a#

#-11 = a#

#a = -11#

The solutions are #a = -11# or #a = 1#