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Sagot :
We have to find the equation of a line that pass through the point (4,3) and is perpendicular to the line:
[tex]y+3=-\frac{9}{11}(x+4)[/tex]This line, that is perpendicular, has a slope m=-9/11, as it is written in slope-point form.
The line we are looking for will have a slope that is the negative reciprocal of m=-9/11, so it will be:
[tex]m_2=-\frac{1}{m_1}=-\frac{1}{-\frac{9}{11}}=\frac{11}{9}[/tex]Our line has a slope of m=11/9.
We can use the known point (4,3) to write the equation as:
[tex]y-3=\frac{11}{9}(x-4)[/tex]If we want the equation in slope intercept form, we rearrange:
[tex]\begin{gathered} y=\frac{11}{9}(x+4)+3 \\ y=\frac{11}{9}x+\frac{44}{9}+3 \\ y=\frac{11}{9}x+\frac{44+27}{9} \\ y=\frac{11}{9}x+\frac{71}{9} \end{gathered}[/tex]Answer: y-3=11/9*(x-4)
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