# What is the equation of the line that is perpendicular to the line passing through (5,3) and (8,8) at midpoint of the two points?

The equation of the line is $5 \cdot y + 3 \cdot x = 47$
The co-ordinates of the mid -point is $\left[\frac{8 + 5}{2} , \frac{8 + 3}{2}\right]$ or $\left(\frac{13}{2} , \frac{11}{2}\right)$ ; The slope m1 of the line passing through $\left(5 , 3\right) \mathmr{and} \left(8 , 8\right)$ is $\frac{8 - 3}{8 - 5}$ or$\frac{5}{3}$; We know the cond ition of perpendicularity of two lines is as $m 1 \cdot m 2 = - 1$ where m1 and m2 are the slopes of the perpendicular lines. So the slope of the line will be $\left(- \frac{1}{\frac{5}{3}}\right)$ or $- \frac{3}{5}$ Now the equation of line passing through the mid point is $\left(\frac{13}{2} , \frac{11}{2}\right)$ is $y - \frac{11}{2} = - \frac{3}{5} \left(x - \frac{13}{2}\right)$ or $y = - \frac{3}{5} \cdot x + \frac{39}{10} + \frac{11}{2}$ or $y + \frac{3}{5} \cdot x = \frac{47}{5}$ or $5 \cdot y + 3 \cdot x = 47$[Answer]