Question

10.If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?

(1 Point)​

Answer 1

Given :–

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

To Find :–

• The possible value of k.

Solution :–

Let, the points be AB.

Now, using distance formula AB,

• $\blue {{\sqrt{{( x_{2} -x_{1}}) ^{2} + {( y_{2} -y_{1})} ^{2} }}}$

Here,

• $x_{1} = 4$
• $x_{2} = 1$
• $y_{1} = k$
• $y_{2} = 0$

Now, put all the values in the distance formula.

→ $\sqrt{ {(1 - 4)}^{2} + {(0 - k)}^{2} } = 5$

Now, solve both brackets.

→ $\sqrt{{( - 3)}^{2} + {( - k)}^{2} } = 5$

Now, open both brackets.

→ $\sqrt{9 + {k}^{2} } = 5$

→ $9 + {k}^{2} = {(5)}^{2}$

→ $9 + {k}^{2} = 25$

→ ${k}^{2} = 25 - 9$

→ ${k}^{2} = 16$

→ $k = \sqrt{16}$

We know that,

• √16 = 4.

So,

→ $k = 4$

Hence,

The value of k is 4.

Check :–

The distance between both points = 5

By using distance formula,

• ${\sqrt{{( x_{2} -x_{1}}) ^{2} + {( y_{2} -y_{1})} ^{2} }}$

• $x_{1} = 4$
• $x_{2} = 1$
• $y_{1} = 4$
• $y_{2} = 0$

Now, put all the values in the distance formula.

→ $\sqrt{ {(1 - 4)}^{2} + {(0 - 4)}^{2} } = 5$

→ $\sqrt{ {(- 3)}^{2} + {(-4)}^{2} } = 5$

→ $\sqrt{9 + 16} = 5$

→ $\sqrt{25} = 5$

We know that,

• √25 = 5.

So,

→ $5 = 5$

Since, the distance between both points is 5.

So, the value of k = 4 is correct.

Answer 2

Step-by-step explanation:

We know that :

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

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

Given : Points are (4 , k) and (1 , 0)

where : x₁ = 4 and x₂ = 1 and y₁ = k and y₂ = 0

Given : Distance between the points (4 , k) and (1 , 0) is 5

Substituting the given values in the Distance formula, We get :

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

Squaring on both sides, We get :

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

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

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

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

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

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

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

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

Answer : Possible values of k are ± 4