The point-slope form of the equation of a straight line is given by the equation:
![\begin{gathered} y-y_1=m(x-x_1) \\ \text{where}\colon \\ (x_1,y_1)\text{ is one point on the line} \\ m\text{ is the slope of the line} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u92i2z8ae72lj6c2upi2d464trt1s2xp46.png)
From the given points:
![\begin{gathered} (0,2)\text{ and (3,-5)} \\ x_1=0;y_1=2 \\ x_2=3;y_2=-5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pujb4nap359f2wbaecchalcjvsmlbf0b1s.png)
We have obtain the slope, m, of the line first. The slope is given by the equation:
![\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(-5-2)/(3-0) \\ m=(-7)/(3) \\ m=(-7)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/78x4t0l7439e7up5i42e9ipcg13yacbi5v.png)
Hence, the point-slope form is:
![\begin{gathered} y-2=-(7)/(3)(x-0)_{} \\ y-2=-(7)/(3)x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dndsv1obsv75b3pnffiks70t6aakoo02am.png)
The slope intercept form of a straight line equation is given by the equation:
![\begin{gathered} y=mx+c \\ \text{where:} \\ m\colon\text{slope} \\ c\colon\text{intercept} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1i7wsgde9hsgkdpysppvaic6g1ilhvvpjs.png)
From the point-slope form, we can deduce the slope-intercept form of the equation.
Thus, we have:
![\begin{gathered} y-2=-(7)/(3)x \\ y=-(7)/(3)x+2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u9qpewe323mvgic7z5bh967dv7jzjw697g.png)
Hence, the slope-intercept form is:
![y=-(7)/(3)x+2](https://img.qammunity.org/2023/formulas/mathematics/college/5717uqmdc9h24f8mf75qy59u2eyhxlh532.png)