Question

which of these points are $sqrt{80}$ units away from (1, 4)? select all that apply. (-7, 0) (9, 7) (-3, -4) (5, -3)

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here $(x_1,y_1)=(1,4)$ and $d = \sqrt{80}$.

Step2: Check point $(-7,0)$

Substitute $x_2=-7,y_2 = 0$ into the distance formula: $d=\sqrt{(-7 - 1)^2+(0 - 4)^2}=\sqrt{(-8)^2+(-4)^2}=\sqrt{64 + 16}=\sqrt{80}$.

Step3: Check point $(9,7)$

Substitute $x_2=9,y_2 = 7$ into the distance formula: $d=\sqrt{(9 - 1)^2+(7 - 4)^2}=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}
eq\sqrt{80}$.

Step4: Check point $(-3,-4)$

Substitute $x_2=-3,y_2=-4$ into the distance formula: $d=\sqrt{(-3 - 1)^2+(-4 - 4)^2}=\sqrt{(-4)^2+(-8)^2}=\sqrt{16 + 64}=\sqrt{80}$.

Step5: Check point $(5,-3)$

Substitute $x_2=5,y_2=-3$ into the distance formula: $d=\sqrt{(5 - 1)^2+(-3 - 4)^2}=\sqrt{4^2+(-7)^2}=\sqrt{16 + 49}=\sqrt{65}
eq\sqrt{80}$.

Answer:

$(-7,0),(-3,-4)$