the value of k if the point p(2,4) is equidistant from the point A(k+1,3) and B(3,2k-1) be
Answers
Answer:
The value of k is equa to 4 or, 2.
The distance AB is 4\sqrt{2}
2
units or, 2\sqrt{13}2
13
units.
Step-by-step explanation:
Given,
The point P(3, 4) is equidistant from the points A(- 2, 3) and B(k, - 1).
To find, the value of k = ? and the distance AB = ?
∴ PA = PB
Using distance formula,
\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
PA = \sqrt{(3+2)^2+(4-3)^2}
(3+2)
2
+(4−3)
2
= \sqrt{(5)^2+(1)^2}
(5)
2
+(1)
2
= \sqrt{26}
26
PB = \sqrt{(k-3)^2+(-1-4)^2}
(k−3)
2
+(−1−4)
2
= \sqrt{k^2-6k+9+25}
k
2
−6k+9+25
= \sqrt{k^2-6k+34}
k
2
−6k+34
\sqrt{k^2-6k+34}
k
2
−6k+34
= \sqrt{26}
26
Squaring both sides, we get
k^2k
2
- 6k + 34 = 26
⇒ k^2k
2
- 6k + 8 = 0
⇒ k^2k
2
- 4k - 2k + 8 = 0
⇒ k(k - 4) -2(k - 4)
⇒ (k - 4)(k - 2) = 0
⇒ k = 4 or, 2
∴ The distance AB =
Using distance formula,
\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Put k = 2
= \sqrt{(2+2)^2+(-1-3)^2}
(2+2)
2
+(−1−3)
2
= \sqrt{16+16}
16+16
= \sqrt{32}
32
= 4\sqrt{2}
2
units
Put k = 4
= \sqrt{(4+2)^2+(-1-3)^2}
(4+2)
2
+(−1−3)
2
= \sqrt{(6)^2+(-4)^2}
(6)
2
+(−4)
2
= \sqrt{36+16}=\sqrt{52}
36+16
=
52
= 2\sqrt{13}2
13
units
Thus, The value of k is equa to 4 or, 2.
The distance AB is 4\sqrt{2}
2
units or, 2\sqrt{13}2
13
units.
