Sunday, February 12, 2012

Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 36

Determine an equation of the line that satisfies the condition "through $(12,10)$; slope $1$".

(a) Write the equation in standard form.

Use the Point Slope Form of the equation of a line with $(x_1,y_1) = (12,10)$ and $m = 1$


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
&& \text{Point Slope Form}
\\
y - 10 =& 1 (x - 12)
&& \text{Substitute $x = 12, y = 10$ and $m=1$}
\\
y - 10 =& x - 12
&& \text{Distributive Property}
\\
-x + y =& -12 + 10
&& \text{Subtract each side by $(x-10)$}
\\
-x + y =& -2
&& \text{Standard Form}
\\
\text{or} &
&&
\\
x - y =& 2
&&

\end{aligned}
\end{equation}
$



(b) Write the equation in slope-intercept form.


$
\begin{equation}
\begin{aligned}

-x + y =& -2
&& \text{Standard Form}
\\
y =& x - 2
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$

No comments:

Post a Comment