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Sagot :
[tex]y=-6x-16[/tex]
Explanation
the slope-intercept form of a line has the form:
[tex]\begin{gathered} y=mx+b \\ where\text{ m is the slope} \\ and\text{ b is the y-intercept} \end{gathered}[/tex]when given the slope and a point of the line we can use the slope-point formula, it says.
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ where\text{ m is the slope} \\ (x_1,y_1)\text{ is a point of the line} \end{gathered}[/tex]so
Step 1
a)Let
[tex]\begin{gathered} slope=\text{ -6} \\ point\text{ \lparen -3,2\rparen} \end{gathered}[/tex]b) now, replace and solve for y
[tex]\begin{gathered} y-y_{1}=m(x-x_{1}) \\ y-2=-6(x-(-3)) \\ y-2=-6(x+3) \\ y-2=-6x-18 \\ add\text{ 2 in both sides} \\ y-2+2=-6x-18+2 \\ y=-6x-16 \end{gathered}[/tex]so, the equation of the line is
[tex]y=-6x-16[/tex]I hope this helps you
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