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Solution
Given triangle ABC with A(3,8), B(7,5), and C(2,3).
Draw the triangle
Calculate the slope of BC
[tex]\begin{gathered} slope,\text{ }m=\frac{change\text{ in y}}{change\text{ in x}} \\ m=\frac{5-3}{7-2} \\ m=\frac{2}{5} \end{gathered}[/tex]Since the line is perpendicular to BC, then the product of the line and line BC = -1
[tex]\begin{gathered} That\text{ is, for perpendicular lines, m}_1m_2=-1 \\ let\text{ m}_2\text{ the slope of the perpendicular line} \\ m_2\text{ x }\frac{2}{5}=-1 \\ m_2=-\frac{1}{\frac{2}{5}} \\ m_2=-\frac{5}{2} \end{gathered}[/tex]The line passes through point A(3,8)
[tex]\begin{gathered} The\text{ equation of the line can be calculated by the formula} \\ y-y_1=m_2(x-x_1) \\ (x_1,y_1)=(3,8) \\ Thus,\text{ y-8}=-\frac{5}{2}(x-3) \\ 2y-16=-5(x-3) \\ 2y-16=-5x+15 \\ 2y=-5x+15+16 \\ 2y=-5x+31 \\ Divide\text{ through by 2} \\ y=-\frac{5}{2}x+\frac{31}{2} \end{gathered}[/tex]Therefore the required equation is:
[tex]y=-\frac{5}{2}x+\frac{31}{2}[/tex]