Solution:
Given the points below;
[tex]\left(-2,3\right)\text{ }and\text{ }(1,4)[/tex]To find the equation of a straight line, the formula is
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]Where
[tex]\begin{gathered} (x_1,y_1)=(-2,3) \\ (x_2,y_2)=(1,4) \end{gathered}[/tex]Substitute the values of the coordinates into the formula to find the equation of a straight line above
[tex]\begin{gathered} \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \frac{y-3}{x-(-2)}=\frac{4-3}{1-(-2)} \\ \frac{y-3}{x+2}=\frac{1}{1+2} \\ \frac{y-3}{x+2}=\frac{1}{3} \\ Crossmultiply \\ 3(y-3)=1(x+2) \\ 3y-9=x+2 \\ x+2=3y-9 \\ x+2-(3y-9)=0 \\ x+2-3y+9=0 \\ x-3y+2+9=0 \\ x-3y+11=0 \end{gathered}[/tex]Hence, the general equation of the line is
[tex]x-3y+11=0[/tex]