Question

Find the distance (-4,5) and (7,18)

Answer 1

Answer:

$\sqrt{290}\ units$

Step-by-step explanation:

The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is

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

Hence, the distance between points $(-4,5)$ and $(7,18)$ is

$\sqrt{(-4-7)^2+(5-18)^2}\\ \\=\sqrt{(-11)^2+(-13)^2}\\ \\=\sqrt{121+169}\\ \\=\sqrt{290}\ units$