Question

Find the value of k, if the point p(2,4) is equidistant from the points a(5, k) and b(k, 7).

Answer 1

given, p(2,4) is equidistant from the points a(5,k) and b(k,7)

let x1=5,x2=k,and x=2

y1=k,y2=7andy=4

since p is the mid point of 'a' and 'b'

x=x1+x2/2 ; y=y1+y2/2

2=5+k/2 ; 4=k+7/2

k=-5 ; 2-7=k

k=-5

therefore, k='-5'

Answer 2

Step-by-step explanation:

Point P is equidistant from Point A and Poimt B.

i.e. Distance AP = Distance BP

Distance formula:

$\sqrt{ {((x2) - (x1))}^{2} + {(y(2) - y(1))}^{2} }$

A = (5 , k)

B = (k , 7)

P (2 , 4)

Given,

Distance AP = Distance BP

$\sqrt{ {(2 - 5)}^{2} + {(4 - k)}^{2} } = \sqrt{ {(2 - k)}^{2} + {(4 - 7)}^{2} } \\ = \sqrt{ {( - 3)}^{2} + {(4 - k)}^{2} } = \sqrt{ {(2 - k)}^{2} + { (- 3)}^{2} } \\ = \sqrt{9 + {(4 - k)}^{2} } = \sqrt{ {(2 - k)}^{2} + 9}$

Squaring both sides,

9 + (4 - k)² = (2 - k)² + 9

=> (4 - k)² = (2 - k)²

=> 16 - 8k + k² = 4 - 4k + k²

=> 16 - 4 = 8k - 4k

=> 12 = 4k

=> k = 3