Determine the value of r such that the line through (5, r) and (2, 3) is perpendicular to the line y = 2x ­- 4 r = 1.5r = 9r = ­- 3r = 4.5

so, what's the slope of y = 2x - 4?  well, is in slope-intercept form, thus

[tex]\bf y=\stackrel{slope}{2}x\stackrel{y-intercept}{-4}[/tex] , so is 2.

now, a perpendicular line to that, will have a negative reciprocal slope to it, thus

[tex]\bf \textit{perpendicular, negative-reciprocal slope for}\quad 2\implies \cfrac{2}{1}\\\\ negative\implies -\cfrac{2}{ 1}\qquad reciprocal\implies - \cfrac{ 1}{2}[/tex]

so, that's its slope of that perpendicular line.

but we also know that it runs through (5,r) and (2,3), so that slope is equal to -1/2

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 5 &,& r~) % (c,d) &&(~ 2 &,& 3~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-r}{2-5}\implies \qquad \qquad \cfrac{3-r}{-3}~~=~~-\cfrac{1}{2} \\\\\\ 3-r=\cfrac{3}{2}\implies 3-\cfrac{3}{2}=r\implies \cfrac{6-3}{2}=r\implies \cfrac{3}{2}=r\implies 1.5=r[/tex]