Question

The distance between (3,k) and ( 5,6) is 2root2 units then k

Answer 1

The points are (3,k) and (5,6).

Given, distance = 2√2 units

Applying distance formula,

Distance = √(x2-x1)^2 + (y2-y1)^2

2√2 = √(x2-x1)^2 + (y2-y1)^2

Squaring both sides,

(2√2)^2 = [√(x2-x1)^2 + (y2-y1)^2]^2

8 = (x2-x1)^2 + (y2-y1)^2

8 = (5-3)^2 + (6-k)^2

8 = 2^2 + (6^2 - 2x6xk + k^2)

8 = 4 + 36 - 12k + k^2

8 = 40 - 12k + k^2

= k^2 - 12k + 32 = 0

On factorising,

=> k^2 - 4k - 8k + 32 = 0

=> k(k-4) -8(k-4) = 0

=> (k-4)(k-8) = 0

=> k-4 = 0 , k-8 = 0

=> k = 4 , k = 8

Answer 2

Explanation:

`√(x) Distance=\sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}\\\\\sqrt{(5-3)^(2)+(6-k)^(2)}=2√(2)\\\\\sqrt{(2)^(2)+[(6)^(2)-(2*6*k)+(k)^(2)]}=2√(2)\\\\\sqrt{4+[36-12k+k^(2)]}=2√(2)\\\\\sqrt{40-12k+k^(2)}=2√(2)\\`

Take square both side

`40-12k+k^(2)=(2√(2))^(2)\\\\40-12k+k^(2)=4*2\\\\40-12k+k^(2)=8\\\\k^(2)-12k+40-8=0\\\\k^(2)-12k+32=0\\`

Factorize,

Sum = -12

Product = 32

Factors = -8 , -4

k² - 8k - 4k + (-8)*(-4) = 0

k(k - 8) - 4(k - 8) = 0

(k - 8)(k - 4) = 0

k = 8 or 4