Question

find the y-coordinates of the points that are 10 units away from the point $(-7,6)$ that have an x-coordinate of 1

the y-coordinates are
(simplify your answer. use a comma to separate answers as needed.)

Explanation:

Step1: Recall distance formula

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

Step2: Plug in known values

We know $d=10$, $(x_1,y_1)=(-7,6)$, $x_2=1$. Let $y_2=y$.
$$10=\sqrt{(1-(-7))^2+(y-6)^2}$$

Step3: Simplify the x-term

Calculate $(1-(-7))=8$, so:
$$10=\sqrt{8^2+(y-6)^2}$$
$$10=\sqrt{64+(y-6)^2}$$

Step4: Square both sides

Eliminate the square root:
$$10^2 = 64+(y-6)^2$$
$$100=64+(y-6)^2$$

Step5: Isolate the squared term

Subtract 64 from both sides:
$$(y-6)^2=100-64=36$$

Step6: Solve for y

Take square roots of both sides:
$$y-6=\pm\sqrt{36}=\pm6$$
Case 1: $y-6=6 \implies y=6+6=12$
Case 2: $y-6=-6 \implies y=6-6=0$

Answer:

0, 12