Question

Which of the following points are more than 5 units from the point P(-2, -2)?

Answer 1

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

$A(x_{1},y_{1}) \\B(x_{2},y_{2}) \\\\D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

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

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(1-(-2))^{2}+(2-(-2)^{2} }\\D=\sqrt{(3)^{2}+(4)^{2} }\\\\D=\sqrt{9+16} \\D=\sqrt{25} \\D=5$

so, A is not more than 5 units from P

Step 2

P(-2,-2)

B(3,-1)

hence

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(3-(-2))^{2}+(-1-(-2)^{2} }\\D=\sqrt{(5)^{2}+(1)^{2} }\\\\D=\sqrt{25+1} \\D=\sqrt{26} \\D=5.099$

so, B is  more than 5 units from P

Step 3

P(-2,-2)

C(2,-3)

hence

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(2-(-2))^{2}+(-3-(-2)^{2} }\\D=\sqrt{(4)^{2}+(-1)^{2} }\\\\D=\sqrt{16+1} \\D=\sqrt{17} \\D=4.123$

so, C is  not more than 5 units from P

Step 4

P(-2,-2)

D(-6,-6)

hence

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(-6-(-2))^{2}+(-6-(-2)^{2} }\\D=\sqrt{(-4)^{2}+(-4)^{2} }\\\\D=\sqrt{16+16} \\D=\sqrt{32} \\D=5.65$

so, D is   more than 5 units from P

Step 5

P(-2,-2)

E(-4,1)

hence

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} } \\D=\sqrt{(-4-(-2))^{2}+(1-(-2)^{2} }\\D=\sqrt{(-2)^{2}+(3)^{2} }\\\\D=\sqrt{4+9} \\D=\sqrt{13} \\D=3.60$

so, E is not  more than 5 units from P

Have a great day.

Answer 2

Answer:D

Step-by-step explanation: