Question

If the distance between the points (4,k) and (1,0) is 5 tgen what can be the possible value of k

Answer 1

4k-10=5
4k=5+10
4k=15
k=15/4

Answer 2

Given:-

• Points are (4 , k) and (1 , 0)
• Distance between the points (4 , k) and (1 , 0) is 5

To Find :-

• Possible value of K

Solution :-

We know that :

Distance between two points (x₁ , y₁) and (x₂ , y₂) is given by :-

$\bigstar\;\;\pink{\mathsf{Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}$

We are given, points are (4 , k) and (1 , 0).Where :-

• x₁ = 4
• x₂ = 1
• y₁ = k
• y₂ = 0

⠀⠀⠀⠀⠀$\small\underline{\pmb{\sf According \: to \: the \: question :-}}$

$\: \: \: \: \: \: \: \:\pink{ :\implies \mathsf{\sqrt{(1 - 4)^2 + (0 - k)^2} = 5}}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{(1 - 4)^2 + (0 - k)^2 = 25}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{(-3)^2 + (-k)^2 = 25}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{9 + k^2 = 25}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{k^2 = 25 - 9}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{k^2 = 16}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{k = \sqrt{16}}$

$\: \: \: \: \: \: \: \: :\implies \mathsf{k = \sqrt{(\pm\;4)^2}}$

$\: \: \: \: \: \: \:\pink{ \: :\implies \mathsf{k = \pm\;4}}$

$\therefore\:\underline{\textsf{ Possible values of k are \textbf{± 4}}}.$