Method 1.
Use the formula of a distance between two points:
[tex]A(x_A;\ y_A),\ B(x_B;\ y_B)\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]
We have
[tex]A(-4,\ 5)\to x_A=-4,\ y_A=5\\B(2,\ -3)\to x_B=2,\ y_B=-3[/tex]
substitute
[tex]|AB|=\sqrt{(-3-5)^2+(2-(-4))^2}=\sqrt{(-8)^2+6^2}=\sqrt{64+36}=\sqrt{100}=10[/tex]
Answer: 10
Method 2
Look at the picture.
Use the Pythagorean theorem:
[tex]x^2=6^2+8^2\\\\x^2=36+64\\\\x^2=100\to x=\sqrt{100}\\\\x=10[/tex]
Answer: 10
