Math Problem Solver

Standard Form:

\[ \begin{aligned}Ax + By + C = 0\end{aligned} \] \[ \begin{aligned} \text{Slope} &= -\frac{A}{B} \\ &= \frac{y_2-y_1}{x_2-x_1} \\ -\frac{A}{B} &= \frac{ 8 - 12 }{ 14 - 5 } \\ -\frac{A}{B} &= \frac{ -4 }{ 9 } \\ A =& 4, B = 9 \end{aligned} \] \[ \begin{aligned} 4 x + 9 y + C &= 0 \\ 4(5) + 9(12) + C &= 0 \\ + + C &= 0 \\ 108 + C &= -20 \\ C &= 128 \end{aligned} \] An equation of the line in standard form is: \[ 4 x + 9 y - 128 = 0 \]

Slope-Intercept Form:

\[ y = m x + b \] \[ \begin{aligned} \text{Slope} &= \frac{y_2-y_1}{x_2-x_1} \\ \text{Slope} &= \frac{ 8 - 12 }{ 14 - 5 } \\ \text{Slope} &= - \frac{4}{9} \end{aligned} \] \[ \begin{aligned} y &= - \frac{4}{9} x + b \\ 12 &= - \frac{4}{9} \times 5 + b \\ 12 &= - \frac{20}{9} + b \\ b &= \frac{128}{9} \end{aligned} \] The slope-intercept form of the line is: \[ y = - \frac{4 x}{9} + \frac{128}{9} \]