If the point P(OK) is equidistant
From the points A (2,-2) and B (6,2)
find K,
Answers
Step-by-step explanation:
The value of k is \frac{15}{8}
8
15
and the length of ap is \frac{5\sqrt{41}}{8}
8
5
41
units.
Step-by-step explanation:
It is given that point p (2,2) is equidistant from the point a(-2,k) and b(-k,3).
Distance formula:
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
pa=pbpa=pb
\sqrt{(-2-2)^2+(k-2)^2}=\sqrt{(-k-2)^2+(3-2)^2}
(−2−2)
2
+(k−2)
2
=
(−k−2)
2
+(3−2)
2
\sqrt{k^2 - 4 k + 20}=\sqrt{k^2 + 4 k + 5}
k
2
−4k+20
=
k
2
+4k+5
Squaring both the sides.
k^2 - 4 k + 20=k^2 + 4 k + 5k
2
−4k+20=k
2
+4k+5
15=8k15=8k
k=\frac{15}{8}k=
8
15
The value of k is \frac{15}{8}
8
15
.
The length of ap is
ap=\sqrt{(-2-2)^2+(k-2)^2}ap=
(−2−2)
2
+(k−2)
2
ap=\sqrt{(-4)^2+(\frac{15}{8}-2)^2}ap=
(−4)
2
+(
8
15
−2)
2
ap=\sqrt{16+(\frac{-1}{8})^2}ap=
16+(
8
−1
)
2
ap=\frac{5\sqrt{41}}{8}ap=
8
5
41
Therefore the length of ap is \frac{5\sqrt{41}}{8}
8
5
41
units.
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