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Sagot :
To write the equation of a paerpendicular line that cross a given point we first need the slope of the given line, then we transform into the spole of the perpendicular line and find the intercept using the given point.
So, we want an equation like this:
[tex]y=ax+b[/tex]And we need "a" and "b". First, let's rewrite the given equation in the slope-interscept form:
[tex]\begin{gathered} 9(2x-4)-6(2y-3)=4y+2 \\ 18x-36-12y+18=4y+2 \\ 18x-12y-18=4y+2 \\ 18x-18-2=4y+12y \\ 18x-20=16y \\ y=\frac{18}{16}x-\frac{20}{16} \\ y=\frac{9}{8}x-\frac{5}{4} \end{gathered}[/tex]This is equivalent to the given graph. 9/8 is the slope. To get the slope of the perpendicular line, we invert it and change its sign. So "a" (the slope of the perpendicular line) is:
[tex]a=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}[/tex]Now we got:
[tex]y=-\frac{8}{9}x+b[/tex]To find "b", we input the values of the point we want it to contain, which is (-81,17):
[tex]17=-\frac{8}{9}(-81)+b[/tex]And we solve for b:
[tex]\begin{gathered} 17=\frac{8\cdot81}{9}+b \\ 17=8\cdot9+b \\ b=17-8\cdot9 \\ b=17-72 \\ b=-55 \end{gathered}[/tex]So, the equation is:
[tex]y=-\frac{8}{9}x-55[/tex]
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