Question

Consider line AB as shown in the graph. On a coordinate plane, a line passes through points (negative 3, 5), (0, 4) and (3, 3). Part A Determine the equation in slope-intercept form that defines line AB : Question 2 Part B Determine the equation in standard form that defines line AB : Question 3 Part C Explain how you determined your equations. Type in the box or use paper to show your work.

Answer 1

to get the equation of any straight line, we simply need two points off of it, hmmm let's use (-3 , 5) and (0 , 4)

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{5}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-3)}}} \implies \cfrac{-1}{0 +3} \implies \cfrac{ -1 }{ 3 } \implies - \cfrac{ 1 }{ 3 }`

`\begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{- \cfrac{ 1 }{ 3 }}(x-\stackrel{x_1}{(-3)}) \implies y -5 = - \cfrac{ 1 }{ 3 } ( x +3)`

`y-5=- \cfrac{ 1 }{ 3 }x-1\implies y=- \cfrac{ 1 }{ 3 }x+4\impliedby \begin{array}c \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}`

now, for the standard form

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

`y=- \cfrac{ 1 }{ 3 }x+4\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y)=3\left( - \cfrac{ 1 }{ 3 }x+4 \right)} \\\\\\ 3y=-x+12\implies x+3y=12\impliedby \textit{standard form}`