How do you find a possible value for a if the points (-2,a), (6,1)has a distance of d=4√5?
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The formula for calculating the distance between two points is:
d
=
√
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
Now, substitute
d
and the values from the points given in the problem and solve for
a
:
7
=
√
(
a
−
−
9
)
2
+
(
5
−
−
2
)
2
7
=
√
(
a
+
9
)
2
+
(
5
+
2
)
2
7
=
√
(
a
+
9
)
2
+
(
7
)
2
7
=
√
(
a
+
9
)
2
+
49
7
2
=
(
√
(
a
+
9
)
2
+
49
)
2
49
=
(
a
+
9
)
2
+
49
49
−
49
=
(
a
+
9
)
2
+
49
−
49
0
=
(
a
+
9
)
2
+
0
0
=
(
a
+
9
)
2
0
=
(
a
+
9
)
(
a
+
9
)
We can now solve
a
+
9
for
0
:
a
+
9
=
0
a
+
9
−
9
=
0
−
9
a
+
0
=
−
9
a
=
−
9
−
9
is a possible value for
a
.
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