Question

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

Answer 1

Answer:

Step-by-step explanation:

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

Take square both side

$40-12k+k^{2}=(2\sqrt{2})^{2}\\\\40-12k+k^{2}=4*2\\\\40-12k+k^{2}=8\\\\k^{2}-12k+40-8=0\\\\k^{2}-12k+32=0$

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

Answer 2

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