If the distance between the points $A (4, k)$ and $B (1, 0)$ is $5 u n i t s$ , then what can be the possible values of $k$ ?

Open in App

Solution

Using distance formula $A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
From the question
$5 = \sqrt{(1 - 4)^{2} + (0 - k)^{2}}$
$\Rightarrow 25 = 9 + k^{2}$ on squaring
$\Rightarrow k^{2} = 25 - 9 = 16$
$∴ k = \pm 4$
$∴ k$ takes $2$ values ${- 4, + 4}$ .

Suggest Corrections

Similar questions

(k, 2 - 2 k), (1 - k, 2 k)

(- k - 4, 6 - 2 k)

k

(x, 3)

(4, 5)

2 \sqrt{5} u n i t s

x =

a_{1}, a_{2}, a_{3}, a_{4}

b

b + 4 \sum K = 1 a_{K} = 8

b^{2} + 4 \sum K = 1 a_{K}^{2} = 16

b

The distance between the lines 6x – 8y + 20 = 0 and 6x – 8y + k = 0 is 1/5, so what are the possible values of K?

Join BYJU'S Learning Program