Math Problem Solver

Slope-Intercept Form:

$$\begin{aligned} y = mx+b\end{aligned}$$$$\begin{aligned}y &= - \frac{4}{3}x + b \\ 1 &= - \frac{4}{3} \times 9 + b \\ 1 &= -12 + b \\ b &= 13 \end{aligned}$$An equation of the line in slope-intercept form is: $y = - \frac{4 x}{3} + 13$

---

Point-Slope Form:

$$\begin{aligned} y - y_{1} = m(x - x_{1}) \end{aligned}$$$$ \text{where m } = - \frac{4}{3}, \text{ } x_{1} = 9, \text{ and } y_{1} = 1 $$An equation of the line in point-slope form is: $y - 1 = - \frac{4 \left(x - 9\right)}{3}$

---

Standard Form:

$$\begin{aligned}Ax + By + C = 0\end{aligned}$$$$\begin{aligned} \text{Slope} &= -\frac{A}{B} \\ -\frac{A}{B} &= \frac{-4}{3} \\ A = 4, B = 3 \end{aligned}$$$$\begin{aligned}4 x + 3 y + C &= 0 \\ 4(9) + 3(1) + C &= 0 \\ 36 + 3 + C &= 0 \\ 36 + C &= -3 \\ C &= -39 \end{aligned}$$An equation of the line in standard form is: $4 x + 3 y - 39 = 0$