equation of a line with the points (3,1) and (5,-4)
You asked:
MathBot Answer:
Standard Form:
\[ \begin{aligned}Ax + By + C = 0\end{aligned} \] \[ \begin{aligned} \text{Slope} &= -\frac{A}{B} \\ &= \frac{y_2-y_1}{x_2-x_1} \\ -\frac{A}{B} &= \frac{ -4 - 1 }{ 5 - 3 } \\ -\frac{A}{B} &= \frac{ -5 }{ 2 } \\ A =& 5, B = 2 \end{aligned} \] \[ \begin{aligned} 5 x + 2 y + C &= 0 \\ 5(3) + 2(1) + C &= 0 \\ + + C &= 0 \\ 15 + C &= -2 \\ C &= 17 \end{aligned} \] An equation of the line in standard form is: \[ 5 x + 2 y - 17 = 0 \]
Slope-Intercept Form:
\[ y = m x + b \] \[ \begin{aligned} \text{Slope} &= \frac{y_2-y_1}{x_2-x_1} \\ \text{Slope} &= \frac{ -4 - 1 }{ 5 - 3 } \\ \text{Slope} &= - \frac{5}{2} \end{aligned} \] \[ \begin{aligned} y &= - \frac{5}{2} x + b \\ 1 &= - \frac{5}{2} \times 3 + b \\ 1 &= - \frac{15}{2} + b \\ b &= \frac{17}{2} \end{aligned} \] The slope-intercept form of the line is: \[ y = - \frac{5 x}{2} + \frac{17}{2} \]