Answer:
The equation of the line in slope intercept form is;
![y=-x+4](https://img.qammunity.org/2023/formulas/mathematics/college/jpmwgbhynvixenrwx6f3gs8fecncvx1wpb.png)
And the equation of the line in point-slope form is;
![y-1_{}=-1(x-3_{})](https://img.qammunity.org/2023/formulas/mathematics/college/62fq2h6qd2z0fzrj4zc97wx8egae24pj26.png)
Step-by-step explanation:
We want to write a linear equation that passes through the points;
![(3,1)\text{ and }(-2,6)](https://img.qammunity.org/2023/formulas/mathematics/college/itd1jzhyxcyr3epdklf9j211ty5sk30uhe.png)
Firstly, let us find the slope of the line;
![\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(6-1)/(-2-3) \\ m=(5)/(-5) \\ m=-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v3t1q8dxtbwwbo537r3fru4zglbornfia1.png)
So, we can now use the point-slope equation of line to find the equation of the line;
![y-y_1=m(x-x_1)_{}](https://img.qammunity.org/2023/formulas/mathematics/college/hfxghqjk47er32rsbtmy1xzcbmtif35kyd.png)
using the point given;
![(x_1,y_1)=(3,1)](https://img.qammunity.org/2023/formulas/mathematics/college/1paimly588c9451rlxwmtm97pehpkf3raj.png)
And the derived slope, we have;
![\begin{gathered} y-y_1=m(x-x_1)_{} \\ y-1_{}=-1(x-3_{}) \\ y-1=-x+3 \\ y=-x+3+1 \\ y=-x+4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hpd0fkwa6d8hy355spqsysc1wwx21lzwkv.png)
Therefore, the equation of the line in slope intercept form is;
![y=-x+4](https://img.qammunity.org/2023/formulas/mathematics/college/jpmwgbhynvixenrwx6f3gs8fecncvx1wpb.png)
And the equation of the line in point-slope form is;
![y-1_{}=-1(x-3_{})](https://img.qammunity.org/2023/formulas/mathematics/college/62fq2h6qd2z0fzrj4zc97wx8egae24pj26.png)