Find the intercepts for each line given:

$2x+3y=12$

${ \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.