${
\begin{array}{l}
\textrm{The $x$-intercept occurs where $y=0$, }\\
\textrm{so we solve $2x+3(0) = 12$ to find $x=6$.}\\\\
\textrm{The $y$-intercept occurs where $x=0$, }\\
\textrm{so we solve $2(0) + 3y=12$ to find $y=4$.}\\\\
\textrm{Remembering intercepts are points, and should}\\
\textrm{be written as ordered pairs, we have intercepts:}\\\\
(6,0),(0,4)
\end{array}}$
$x-5y=7$
${
\begin{array}{l}
\textrm{The $x$-intercept occurs where $y=0$, }\\
\textrm{so we solve $x-5(0)=7$ to find $x=7$.}\\\\
\textrm{The $y$-intercept occurs where $x=0$, }\\
\textrm{so we solve $(0)-5y=7$ to find $y=-7/5$.}\\\\
\textrm{Remembering intercepts are points, and should}\\
\textrm{be written as ordered pairs, we have intercepts:}\\\\
(7,0),(0,-7/5)
\end{array}}$
$x=7$
${
\begin{array}{l}
\textrm{The line $x=7$ is a vertical line and consequently, does}\\
\textrm{not have a $y$-intercept -- only an $x$-intercept at $(7,0)$}
\end{array}}$
Find the equation in slope-intercept form for the line containing each pair of points given:
$(0,1), (2,5)$
${
\begin{array}{l}
\textrm{First, we find the slope $m = \displaystyle{\frac{5-1}{2-0} = \frac{4}{2} = 2}$}\\\\
\textrm{With one of the two points, say $(0,1)$, we can write the equation for the line}\\
\textrm{with slope } m \textrm{ through that point: $y - 1 = 2(x - 0)$. Solving for } y \textrm{ reveals}\\
\textrm{the slope-intercept form: $y = 2x + 1$.}
\end{array}}$
$(-1,3), (-2,-1)$
${
\begin{array}{l}
\textrm{First, we find the slope $m = \displaystyle{\frac{-1-3}{-2-(-1)} = \frac{-4}{-1} = 4}$}\\\\
\textrm{With one of the two points, say $(-2,-1)$, we can write the equation for the line}\\
\textrm{with slope } m \textrm{ through that point: $y - (-1) = 4(x - (-2))$. Solving for } y \textrm{ reveals}\\
\textrm{the slope-intercept form: $y = 4x + 7$.}
\end{array}}$
$(-3,2)$, $(11,-7)$
${\displaystyle{y=-\frac{9}{14}x+\frac{55}{14}}}$
Find the equation in point-slope form for the line parallel to the line described by $7x-2y=4$ and going through the point $(2,-1)$.
${
\begin{array}{l}
\textrm{Parallel lines share the same slope, so we rewrite $7x-2y=4$}\\
\textrm{in slope-intercept form so that we may identify the slope. }\\\\
\textrm{Doing this, we find $y = \displaystyle{\frac{7}{2} x-2}$, which has slope $\displaystyle{\frac{7}{2}}$.}\\\\
\textrm{Given that the desired line goes through $(2,-1)$, we }\\
\textrm{employ the point-slope form to find it's equation:}\\\\
y+1=\displaystyle{\frac{7}{2}(x-2)}
\end{array}}$
Write the the equations of the lines parallel and perpendicular to $3x+2y=5$ and going through $(3,-1)$ in point-slope form, slope-intercept form, and standard form.
${
\begin{array}{l}
\textrm{Parallel lines share the same slope, so we rewrite $3x+2y=5$}\\
\textrm{in slope-intercept form so that we may identify the slope. }\\\\
\textrm{Doing this, we find $y = \displaystyle{-\frac{3}{2} x-\frac{5}{2}}$, which has slope $\displaystyle{-\frac{3}{2}}$.}\\\\
\textrm{Given that the desired line goes through $(3,-1)$, we }\\
\textrm{employ the point-slope form to find it's equation:}\\\\
y+1=\displaystyle{-\frac{3}{2}(x-3)}\\\\
\textrm{Solving for $y$ reveals the slope-intercept form:}\\\\
y = \displaystyle{-\frac{3}{2} x +\frac{7}{2}}\\\\
\textrm{Multiplying by $2$ to clear the fractions, and putting the $x$ and }\\
\textrm{$y$ terms on the left side reveals the standard form:}\\\\
3x + 2y = 7\\\\
\textrm{Perpendicular lines have slopes that are negative reciprocals,}\\
\textrm{so the slope of the line perpendicular to the given one is $2/3$.}\\
\textrm{Here again, we can use this in combination with the given}\\
\textrm{point $(3,-1)$ to find the point-slope form for this line:}\\\\
y+1 = \frac{2}{3} (x-3)\\\\
\textrm{The slope-intercept and standard forms of this same line}\\
\textrm{are produced in a way similar to what was used for the }\\
\textrm{parallel line:}\\\\
y= \displaystyle{\frac{2}{3} x - 3} \quad \quad \textrm{(slope-intercept form)}\\
2x-3y=9 \quad \quad \textrm{(standard form)}
\end{array}}$
Find the slope-intercept form for a line that passes through a point $(x_0,y_0)$ with slope $m$ without using the point-slope form for a line. Show that the equation that results is equivalent to the point-slope form for this line.
