Respuesta :
Answer:
B(3,-1),D.(-6,-6)
Step-by-step explanation:
Hello
the distance between two points is equal to the length of the line segment that joins them, expressed numerically.
Let 2 points
[tex]A(x_{1},y_{1}) \\B(x_{2},y_{2}) \\\\D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }[/tex]
all you have to do is to find the distance between P and each point replacing the values.
Step 1
P(-2,-2)
A(1,2)
hence
[tex]D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(1-(-2))^{2}+(2-(-2)^{2} }\\D=\sqrt{(3)^{2}+(4)^{2} }\\\\D=\sqrt{9+16} \\D=\sqrt{25} \\D=5[/tex]
so, A is not more than 5 units from P
Step 2
P(-2,-2)
B(3,-1)
hence
[tex]D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(3-(-2))^{2}+(-1-(-2)^{2} }\\D=\sqrt{(5)^{2}+(1)^{2} }\\\\D=\sqrt{25+1} \\D=\sqrt{26} \\D=5.099[/tex]
so, B is more than 5 units from P
Step 3
P(-2,-2)
C(2,-3)
hence
[tex]D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(2-(-2))^{2}+(-3-(-2)^{2} }\\D=\sqrt{(4)^{2}+(-1)^{2} }\\\\D=\sqrt{16+1} \\D=\sqrt{17} \\D=4.123[/tex]
so, C is not more than 5 units from P
Step 4
P(-2,-2)
D(-6,-6)
hence
[tex]D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(-6-(-2))^{2}+(-6-(-2)^{2} }\\D=\sqrt{(-4)^{2}+(-4)^{2} }\\\\D=\sqrt{16+16} \\D=\sqrt{32} \\D=5.65[/tex]
so, D is more than 5 units from P
Step 5
P(-2,-2)
E(-4,1)
hence
[tex]D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(-4-(-2))^{2}+(1-(-2)^{2} }\\D=\sqrt{(-2)^{2}+(3)^{2} }\\\\D=\sqrt{4+9} \\D=\sqrt{13} \\D=3.60[/tex]
so, E is not more than 5 units from P
Have a great day.
