Question

pa=pb
find 'k'
p(-2,4), A(4,k), B(k,-2)

Answer 1

Step-by-step explanation:

using distance formula we have

$pa = \sqrt{ {(4 +2) }^{2} + {(k - 4)}^{2} }$

$pb = \sqrt{ {(k+2) }^{2} + {( - 2- 4)}^{2} }$

now,

since pa=pb

therefore,

$\sqrt{ {(4 +2) }^{2} + {(k - 4)}^{2} } = \\ \sqrt{ {(k+2) }^{2} + {( - 2- 4)}^{2} }$

$\\ = > {6}^{2} + {k}^{2} + {4}^{2} - 8k = \\ {k}^{2} + {4}^{2} + 8k + {( - 6)}^{2}$

$\\ = > 36 + {k}^{2} + 16 - 8k = \\ {k}^{2} + 16 + 8k + 36$

$\\ = > 8k = - 8k$

$= > 8k + 8k = 0$

$= > 16k = 0$

$= > k = 0$