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Sagot :
Given:
point (2,-3) and (-3,7)
First, solve for the slope of the line
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \text{substitute} \\ (x_1,y_1)=(2,-3) \\ (x_2,y_2)=(-3,7) \\ \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{7-(-3)}{-3-2} \\ m=\frac{7+3}{-5} \\ m=\frac{10}{-5} \\ m=-2 \end{gathered}[/tex]Now that we have solved for the slope, substitute it to the slope-intercept form y = mx + b to find the y-intercept. Use the point (2,-3) but using (-3,7) works just as well.
[tex]\begin{gathered} y=mx+b \\ -3=(-2)(2)+b \\ -3=-4+b \\ -3+4=b \\ b=1 \end{gathered}[/tex]The slope intercept form of the line is
[tex]y=-2x+1[/tex]Rearrange it to follow the standard form Ax + By = C, and we have
[tex]\begin{gathered} y=-2x+1 \\ y+2x=-2x+2x+1 \\ 2x+y=\cancel{-2x+2x}+1 \\ \\ \text{The equation of the line that passes through (2,-3) and (-3,7) in standard form is} \\ 2x+y=1 \end{gathered}[/tex]
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