If the point p (2,2) is equidistant from the point a(-2,k) and b(-2k,-3), find k. also find the length of ap
Answers
Step-by-step explanation:
The given points are P(2, 2), A(−2, k) and B(−2k, −3).
We know that the distance between the points,(x1,y1) and (x2,y2)is given by:
d
=
√
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
d=(x2-x1)2+(y2-y1)2
It is given that P is equidistant from A and B.
∴ AP = BP
⇒ AP2 = BP2
⇒ (2 − (−2))2 + (2 − k)2 = (2 − (−2k))2 + (2 − (−3))2
⇒ (4)2 + (2 − k)2 = (2 + 2k)2 + (5)2
⇒ 16 + k2 + 4 − 4k = 4 + 4k2 + 8k + 25
⇒ 3k2 + 12k + 9 = 0
⇒ k2 + 4k + 3 = 0
⇒ k2 + 3k + k + 3 = 0
⇒ (k + 1) (k + 3) = 0
⇒ k = −1, −3
Thus, the value of k is −1 and −3.
For k = −1:
Length of AP
=
√
(
2
−
(
−
2
)
)
2
+
(
2
−
1
(
−
1
)
)
2
=
√
4
2
+
3
2
=
√
16
+
9
=
√
25
=
5
=(2-(-2))2+(2-1(-1))2=42+32=16+9=25=5
For k = −3:
Length of AP
=
√
(
2
−
(
−
2
)
)
2
+
(
2
−
1
(
−
3
)
)
2
=
√
4
2
+
5
2
=
√
16
+
25
=
√
41
=(2-(-2))2+(2-1(-3))2=42+52=16+25=41
Thus, the length of AP is either
5
units
5 units or
√
41
units
.
41units.
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Answer:
Step-by-step explanation:
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