[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-7}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[1-(-7)]^2+[y-6]^2}\implies 10^2=[1-(-7)]^2+[y-6]^2 \\\\\\ 100=(1+7)^2+(y-6)(y-6)\implies 100=8^2+\stackrel{F~O~I~L}{(y^2-12y+36)} \\\\\\ 100=64+36+y^2-12y\implies 100=100+y^2-12y \\\\\\ 0=y^2-12y\implies 0=y(y-12)\implies y= \begin{cases} 0\\ 12 \end{cases}[/tex]
For a given point (x₁, y₁) the set of all the points that are at a fixed distance R from the point form a circle, such that the equation of the circle is given by:
(x - x₁)^2 + (y - y₁)^2 = R^2
We will find that the two solutions are:
Here we want to find the y-coordinates of all the points that are 10 units away from the point (-7, 6) with an x-component of 1.
First, the circle will be:
(x + 7)^2 + (y - 6)^2 = 10^2
Now the x component must be equal to 1, so x = 1
(1 + 8)^2 + (y - 6)^2 = 100
81 + (y - 6)^2 = 100
(y - 6)^2 = 100 - 81 = 19
(y - 6)^2 = 19
(y - 6) = ±√19 = ±4.36
y = ±4.36 + 6
Then the two y-coordinates are:
y = 4.36 + 6 = 10.37
y = -4.36 + 6 = 1.64
So the two points are:
(1, 10.37) and (1, 1.64)
If you want to learn more, you can read:
https://brainly.com/question/23988015
