$m = \frac{2 + 5}{-1+4} = \frac{7}{3}$

$m = \frac{-1-(-4)}{-5-1} = \frac{3}{-6} = -\frac{1}{2}$

$m = \frac{5 - 3}{-2 - 1} = -\frac{2}{3}$

$m = \frac{4-3}{0-(-2)} = \frac{1}{2}$

$m = \frac{-1 - 2}{4 + 7} = -\frac{3}{11}$

$m = \frac{-2-7}{-3-1} = \frac{-9}{-4} = \frac{9}{4}$

$m = \frac{3 - 5}{-4 + 1} = \frac{-2}{-3} = \frac{2}{3}$

$m = \frac{4-0}{-1-6} = \frac{4}{-7} = -\frac{4}{7}$

$1/2$

$-2$

$1$

$3$

$y = -1x + 2$, so the slope is $-1$.

$y = 1x + 3$, so the slope is $1$.

$y = -1x + 1$, so the slope is $-1$.

$y = 1x + 5$, so the slope is $1$.

$y = \frac{2}{3}x + 1$, so the slope is $\frac{2}{3}$.

$y = -\frac{1}{2}x - 7$, so the slope is $-\frac{1}{2}$.

\begin{align*} x + 3y - 5 &= 0\\ 3y &= -x + 5\\ y &= -\frac{1}{3}x + \frac{5}{3} \end{align*} so the slope is $-\frac{1}{3}$.

\begin{align*} m_1 &= 2\\ m_2 &= -\frac{1}{m_1}\\ &= -\frac{1}{2}\end{align*}

\begin{align*} m_1 &= -\frac{3}{5}\\ m_2 &= -\frac{1}{m_1}\\ &= \frac{5}{3}\end{align*}

\begin{align*} m_1 &= -4\\ m_2 &= -\frac{1}{m_1}\\ &= \frac{1}{4}\end{align*}

\begin{align*} m_1 &= \frac{1}{7}\\ m_2 &= -\frac{1}{m_1}\\ &= -7\end{align*}

\begin{align*} m_1 &= 6\\ m_2 &= -\frac{1}{m_1}\\ &= -\frac{1}{6}\end{align*}

\begin{align*} m_1 &= -\frac{2}{3}\\ m_2 &= -\frac{1}{m_1}\\ &= \frac{3}{2}\end{align*}

\begin{align*} m_1 &= -2\\ m_2 &= -\frac{1}{m_1}\\ &= \frac{1}{2}\end{align*}

\begin{align*} m_1 &= \frac{1}{3}\\ m_2 &= -\frac{1}{m_1}\\ &= -3\end{align*}

Method one: \begin{align*} y &= 2x + c\\ 3 &= 2(-1) + c\\ c &= 5\\ \therefore y&=2x + 5 \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-3 &= 2(x-(-1))\\ &= 2x + 2\\ y &= 2x + 5 \end{align*}

Method one: \begin{align*} y &= \frac{1}{3}x + c\\ -2 &= \frac{1}{3}(2) + c\\ c &= -2 - \frac{2}{3}\\&= -\frac{8}{3}\\ \therefore y&=\frac{1}{3}x -\frac{8}{3} \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-(-2) &= \frac{1}{3}(x-2)\\ y+2&= \frac{1}{3}x - \frac{2}{3}\\ y &= \frac{1}{3}x - \frac{8}{3} \end{align*}

Method one: \begin{align*} y &= -3x + c\\ -1 &= -3\times4 + c\\ c &= 11\\ \therefore y&=-3x + 11 \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-(-1) &= -3(x-4)\\ y+1 &= -3x + 12\\ y &= -3x + 11 \end{align*}

Method one: \begin{align*} y &= -\frac{1}{2}x + c\\ 1 &= -\frac{1}{2}(-2) + c\\ c &= 0\\ \therefore y&=-\frac{1}{2}x \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-1 &= -\frac{1}{2}(x-(-2))\\ &= -\frac{1}{2}x - 1\\ y &= -\frac{1}{2}x \end{align*}

Method one: \begin{align*} y &= 1x + c\\ \frac{1}{2} &= 5 + c\\ c &= -\frac{9}{2}\\ \therefore y&=x - \frac{9}{2} \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-\frac{1}{2} &= 1(x-5)\\ y &= x - \frac{9}{2} \end{align*}

Method one: \begin{align*} y &= \frac{2}{3}x + c\\ -2 &= \frac{2}{3}(-6) + c\\&= -4 + c\\ c &= 2\\ \therefore y&=\frac{2}{3}x + 2 \end{align*} Method two: \begin{align*} y-y_1 &= m(x-x_1)\\ y-(-2) &= \frac{2}{3}(x-(-6))\\ y+2 &= \frac{2}{3}x + 4\\ y &= \frac{2}{3}x + 2 \end{align*}