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3 years ago

Write an equation of the line that is parallel to y = 1/2x + 3 and passes through the point (10, -5).

Mathematics

1 answer:

bearhunter [10]3 years ago

8 0

y = mx + b

"m" is the slope, "b" is the y-intercept (the y value when x = 0)

You need to find "m" and "b"

For lines to be parallel, the slopes have to be the same.

The given line's slope is 1/2, so the parallel line's slope is also 1/2.

y = 1/2x + b

To find "b", you can plug in the point (10, -5) into the equation.

y = 1/2x + b

-5 = 1/2(10) + b

-5 = 5 + b Subtract 5 on both sides to get "b" by itself

-10 = b

y = 1/2x - 10 Your answer is D (you could have just found the slope to get your answer, since all the slopes are different)

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2 years ago

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Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir

kati45 [8]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}$

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$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
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$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
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(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
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$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
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(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
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