Using two-points form equation of a line formula as shown below
![(y-y_1)/(x-x_1)=(y_2-y__1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/college/1y8w3h20y12qnjagbe831rvxqvorpjw9nv.png)
From the question, we have that:
![\begin{gathered} (1,1)\rightarrow x_1=1,y_1=1 \\ (2,-1)\rightarrow x_2=2,y_2=-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cizya5n5y6salqipvulz18i266ko0mqpft.png)
Substitute the coordinates values above in the two-points form equation
![(y-1)/(x-1)=(-1-1)/(2-1)](https://img.qammunity.org/2023/formulas/mathematics/college/knr7e58edphscumos6jwky6y29xklzc84v.png)
![(y-1)/(x-1)=(-2)/(1)](https://img.qammunity.org/2023/formulas/mathematics/college/88eumpzmpyti91c3x8fh6azezoc26pbhdd.png)
Cross multiply and simplify the equation
![\begin{gathered} 1(y-1)=-2(x-1) \\ y-1=-2x+2 \\ y=-2x+2+1 \\ y=-2x+3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4kqj91ahl7j3m6iuzp8g9ezdu3z7lxsebu.png)
Hence, the equation of the straight line is y = -2x + 3