Question

Find the distance of PQ P(0,4) and Q(10,-6)

Answer 1

Answer:

$PQ=10\sqrt{2}\:\: \sf units$

Step-by-step explanation:

Distance between two points formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}$

Define the variables:

• Let (x₁, y₁) = (0, 4)
• Let (x₂, y₂) = (10, -6)
• d = PQ

Substitute the defined variables into the distance formula and solve for PQ:

$\implies PQ=\sqrt{(10-0)^2+(-6-4)^2}$

$\implies PQ=\sqrt{10^2+(-10)^2}$

$\implies PQ=\sqrt{100+100}$

$\implies PQ=\sqrt{200}$

$\implies PQ=\sqrt{100 \cdot 2}$

$\implies PQ=\sqrt{100}\sqrt{2}$

$\implies PQ=10\sqrt{2}\: \sf units$

Learn more about the distance between two points here:

brainly.com/question/28144723

Answer 2

Answer:

PQ ≈ 14.14 units

Step-by-step explanation:

calculate the distance d using the distance formula

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

with (x₁, y₁ ) = P (0, 4 ) and (x₂, y₂ ) = Q (10, - 6 )

PQ = $\sqrt{(10-0)^2+(-6-4)^2}$

     = $\sqrt{10^2+(-10)^2}$

     = $\sqrt{100+100}$

     = $\sqrt{200}$

     ≈ 14.14 units ( to 2 dec. places )