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Aleks [24]
3 years ago
7

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

Mathematics
1 answer:
prohojiy [21]3 years ago
7 0

\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}

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Answer:

see explanation

Step-by-step explanation:

Note that cos315° = cos45° and sin315° = - sin45° and

cos45° = sin45° = \frac{1}{\sqrt{2} } = \frac{\sqrt{2} }{2}

Hence

12(cos315° + isin315°)

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