Question

find the y - coordinates of the points that are 10 units away from the point (1,6) that have an x - coordinate of - 7.
the y - coordinates are
(simplify your answer. use a comma to separate answers as needed.)

Explanation:

Step1: Recall the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $(x_1,y_1)=(1,6)$ and $(x_2,y_2)=(-7,y)$. We know that $d = 10$.

Step2: Substitute values into the formula

Substitute the values into the distance formula: $10=\sqrt{(-7 - 1)^2+(y - 6)^2}$. First, simplify $(-7 - 1)^2$: $(-7 - 1)^2=(-8)^2 = 64$. So the equation becomes $10=\sqrt{64+(y - 6)^2}$.

Step3: Square both sides

Square both sides of the equation to get rid of the square - root: $100=64+(y - 6)^2$.

Step4: Isolate the squared term

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

Step5: Solve for y

Take the square root of both sides: $y - 6=\pm\sqrt{36}=\pm6$.
When $y - 6 = 6$, then $y=6 + 6=12$.
When $y - 6=-6$, then $y=-6 + 6=0$.

Answer:

0,12