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Sagot :
Given that, t is the perpendicular bisector of CD. Therefore, t passes through the midpoint of CD.
The midpoint of c(-5,9) and D(7,5) is,
[tex]\begin{gathered} \frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \\ E(\frac{-5+7}{2},\frac{9+5}{2}) \\ E(1,7) \end{gathered}[/tex]Therefore, the point (1,7) passes through the line t.
The slope of the line segemnet joing (-5,9) and (7,5) is,
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ =\frac{5-9}{7+5} \\ =-\frac{1}{3} \end{gathered}[/tex]Use the equation m1m2=-1 to calculate the solpe of line t.(Both are perpendicular).
[tex]\begin{gathered} -\frac{1}{3}m_2=-1 \\ m_2=3 \end{gathered}[/tex]Calculate the equation of line t having slope 3 and point (1,7)
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-7=3(x-1) \\ y-7=3x-3 \\ y=3x+4 \end{gathered}[/tex]Therefore, the slope intersept form is y=3x+4.
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