Answer:
d = √10
Step-by-step explanation:
[tex]K(-1, -3) , L(0, 0).\\\\d=\sqrt{((x_2-x_1)^2+ (y_2-y_1)^2) } \\\\x_1 =-1\\\\y_1 =-3\\\\x_2 =0\\\\y_2 =0 \\\\d = \sqrt{(0-(-1))^2+(0-(-3))^2}\\\\ d = \sqrt{(0+1)^2+(0+3)^2}\\\\ d = \sqrt{(1)^2 + (3)^2}\\\\ d = \sqrt{1 + 9}\\\\ d = \sqrt{10} \\[/tex]
Answer:
[tex]\huge\boxed{|KL|=\sqrt{10}\approx3.2}[/tex]
Step-by-step explanation:
The formula of a distance between two points (x₁; y₁) and (x₂; y₂):
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
We have K(-1; -3) and L(0; 0). Substitute:
[tex]|KL|=\sqrt{(0-(-3))^2+(0-(-1))^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}}[/tex]
Look at the picture.
We have the right triangle with the legs 3 and 1.
Use the Pythagorean theorem:
[tex]leg^2+leg^2=hypotenuse^2[/tex]
substitute:
[tex]3^2+1^2=|KL|^2\\\\|KL|^2=9+1\\\\|KL|^2=10\to|KL|=\sqrt{10}[/tex]
