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Sagot :
Given the equation of the line below,
[tex]8y-16=5x[/tex]If the line passes through the point,
[tex](5,-5)[/tex]Re-writing the eqaution of the line in slope intercept form,
[tex]\begin{gathered} 8y-16=5x \\ 8y=5x+16 \\ \text{Divide both sides by 8} \\ y=\frac{5x}{8}+\frac{16}{2} \\ y=\frac{5}{8}x+2 \end{gathered}[/tex]The slope of the perpendicular line is the negative reciprocal of the slope of the eqaution of the line in the slope-intercept form given above
The general form of the slope-intercept form of the equation of a straight line is,
[tex]\begin{gathered} y=mx+c \\ \text{Where m is the slope} \\ y=\frac{5}{8}x+2 \\ m=\frac{5}{8} \\ \text{Slope of the perpendicular line is} \\ m_1=-\frac{1}{m} \\ m_{1_{}}=-\frac{1}{\frac{5}{8}}=-1\times\frac{8}{5}=-\frac{8}{5} \end{gathered}[/tex]The formula to find the equation of a line with point (5, -5) below is,
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=m_1 \\ \text{Where} \\ (x_1,y_1)=(5,-5) \\ m_1=-\frac{8}{5} \end{gathered}[/tex]Substitute the values into the formula of the eqaution of a straight line,
[tex]\begin{gathered} \frac{y-(-5)}{x-5}=-\frac{8}{5} \\ \frac{y+5}{x-5}=-\frac{8}{5} \\ \text{Crossmultiply} \\ 5(y+5)=-8(x-5) \\ 5y+25=-8x+40 \\ \text{Collect like terms} \\ 5y=-8x+40-25 \\ 5y=-8x+15 \\ \text{Divide both sides by 5} \\ \frac{5y}{5}=-\frac{8}{5}x+\frac{15}{5} \\ y=-\frac{8}{5}x+3 \end{gathered}[/tex]Hence, the right option is C
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