Question

Find value of k for which the distance between the points P(1,k) and Q(4,2) is 5 units

Answer 1

Answer:

RESGE

Step-by-step explanation:

Answer 2

       $\underline{\bold{Solution:}}$

       $\text{The two given point are }P(1,k) \text{ and }Q(4,2)$

       $\text{It is also given that }QP = 5 \text{ units}$

       $\text{Applying distance formula:}$

       $\sqrt{(x_1 - x_2)^{2} + (y_1-y_2)^{2}} = \text{Distance between }(x_1,y_1) \text{ and }(x_2,y_2)$

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

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

       $\text{Squaring both sides}$

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

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

$\implies 9 + (k-2)^{2} = 25$

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

$\implies (k-2)^{2} = 16$

       $\text{Rooting both sides}$

$\implies \sqrt{(k-2)^{2}} = \sqrt{16}$

$\implies k-2 = 4$

$\implies k = 4 + 2$

$\implies \boxed{k = 6}$