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Alex
2 years ago
11

[PLEASE ANSWER] Precalculus Question (has 2 parts)

Mathematics
1 answer:
artcher [175]2 years ago
3 0
Part I

\bf \begin{array}{l|lllllll}
x&2&\boxed{4}&6&8&\boxed{10}&12&14\\\\
y&1&\boxed{6}&11&16&\boxed{21}&26&31
\end{array}\implies 
\begin{array}{llll}
(4,6)\\\\
(10,21)
\end{array}\\\\
-------------------------------\\\\

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 4}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 10}}\quad ,&{{ 21}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{21-6}{10-4}


part II

\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=6\\
x_1=4\\
m=\boxed{?}
\end{cases}\\
\end{array}\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}


now, the form above is the point-slope form, and that'd be the equation, you can leave it like so, or you can solve for "y", and put it in slope-intercept, same equation though.
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