Question

Find all the values of k so that the given points are 29 square units apart. (-5, 5) and (k,0).

Answer 1

Distance formula between 2 points (x1,y1) and (x2,y2)

D=$\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}$

D=29 and one point is (-5,5) and the other is (k,0)
sub

29=$\sqrt{(-5-k)^{2}+(5-0)^{2}}$
29=$\sqrt{(-5-k)^{2}+(5)^{2}}$
square both sides
841=$(-5-k)^{2}+(5)^{2}$
841=$(-5-k)^{2}+25$
minus 25 both sides
816=$(-5-k)^{2}$
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k




the possible values for k are
k=-5+4√51 or k=-5-4√51