Find the coordinates of the points that are 13 units away from the origin and have an x-coordinate equal to 12.

given (0,0) and (12,y1)
the distance formula states taht
the distance between the points (x1,y1) and (x2,y2) is

D=[tex] \sqrt{(x2-x1)^2+(y2-y1)^2} [/tex]
given
D=13
x1=0, y1=0
x2=12

[tex]13= \sqrt{(12-0)^2+(y2-0)^2} [/tex]
[tex]13= \sqrt{(12)^2+(y2)^2} [/tex]
[tex]13= \sqrt{144+(y2)^2} [/tex]
square both sides

[tex]169= 144+(y2)^2 [/tex]
minus 144 both sides
[tex]25=(y2)^2 [/tex]
sqrt both sides
+/-5=y2

the points are
(12,5) and (12,-5)

Answer Link

joaobezerra joaobezerra · Answer 2 · 2021-08-10T21:01:01+02:00

This question is solved using the distance between two points formula, and doing this, we get that the coordinates are: (12,5) and (12,-5).

Distance between two points:

Suppose that we have two points, [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]. The distance between them is given by:

[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

In this question:

Origin: Coordinates (0,0).
Point with a x-coordinate of 12: P(12,y)

13 units away from the origin:

This means that the distance between P and the origin is 13, so:

[tex]\sqrt{(12-0)^2+(y-0)^2} = 13[/tex]

[tex](\sqrt{(12-0)^2+(y-0)^2})^2 = 13^2[/tex]

[tex]12^2 + y^2 = 13^2[/tex]

[tex]144 + y^2 = 169[/tex]

[tex]y^2 = 25[/tex]

[tex]y = \pm \sqrt{25}[/tex]

[tex]y = \pm 5[/tex]

Thus, the coordinates are: (12,5) and (12,-5).

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