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If the point P(2, 2) is equidistat from the points A(-2, k) an

Question

If the point $P (2, 2)$ is equidistat from the points $A (- 2, k)$ and $B (- 2 k, - 3)$ , find k. Also, find the length of AP.

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Jack Discussed

If the point

P (2, 2)

is equidistat from the points

A (- 2, k)

and

B (- 2 k, - 3)

, find k. Also, find the length of AP.

11 mins ago

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11 mins ago

Text solutionVerified

Given: Point

P (2, 2)

is equidistant from the two points

A (- 2, k)

and

B (- 2 k, - 3)

P A = PB

P A^{2} = P B^{2}

(2 + 2)^{2} + (2 - k)^{2} = (2 + 2 k)^{2} (2 + 3)^{2}

4^{2} + 4 - 4 k + k^{2} = 4 + 8 k + 4 k^{2} + 5^{2}

16 + 4 - 4 k + k^{2} = 4 + 8 k + 4 k^{2} + 25

4 k^{2} + 8 k + 29 - 20 + 4 k - k^{2} = 0

3 k^{2} + 12 k + 9 = 0

k^{2} + 4 k + 3 = 0

k^{2} + k + 3 k + 3 = 0

k (k + 1) + 3 (k + 1) = 0

(k + 1) (k + 3) = 0

thus,

k = - 1

k = - 3

k = - 1

A P^{2} = 20 - 4 k + k^{2}

= 20 + 4 + 1

= 25

= 5

units

k = - 3

A P^{2} = 20 - 4 k + k^{2}

= 20 + 12 + 9

= 41

A P = 41

units.

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Question Text	If the point $P (2, 2)$ is equidistat from the points $A (- 2, k)$ and $B (- 2 k, - 3)$ , find k. Also, find the length of AP.
Answer Type	Text solution:1
Upvotes	150