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

1 Answer

2 votes
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