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Which point is 7 units from (–2, 4)?

A.
(–5, 4)

B.
(–2, 3)

C.
(5, 4)

D.
(9, 4)

Sagot :

Naǫidn
The distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex](x_1,y_1)=(-2,4) \\ d=7 \\ \\ 7=\sqrt{(x-(-2))^2+(y-4)^2} \\ 7=\sqrt{(x+2)^2+(y-4)^2} \ \ \ |^2 \\ 49=(x+2)^2+(y-4)^2[/tex]
Check which point satisfies the equation:
[tex](x,y)=(-5,4) \\ 49 \stackrel{?}{=} (-5+2)^2+(4-4)^2 \\ 49 \stackrel{?}{=} (-3)^2+0^2 \\ 49 \stackrel{?}{=} 9 \\ 49 \not= 9 \\ doesn't \ satisfy \ the \ equation[/tex]

[tex](x,y)=(-2,3) \\ 49 \stackrel{?}{=} (-2+2)^2+(3-4)^2 \\ 49 \stackrel{?}{=} 0^2+(-1)^2 \\ 49 \stackrel{?}{=} 1 \\ 49 \not= 1 \\ doesn't \ satisfy \ the \ equation[/tex]

[tex](x,y)=(5,4) \\ 49 \stackrel{?}{=} (5+2)^2+(4-4)^2 \\ 49 \stackrel{?}{=} 7^2+0^2 \\ 49 \stackrel{?}{=} 49 \\ 49=49 \\ satisfies \ the \ equation[/tex]

[tex](x,y)=(9,4) \\ 49 \stackrel{?}{=} (9+2)^2+(4-4)^2 \\ 49 \stackrel{?}{=} 11^2+0^2 \\ 49 \stackrel{?}{=} 121 \\ 49 \not= 121 \\ doesn't \ satisfy \ the \ equation[/tex]

The answer is C.
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