5y = -2x + 3
y = -2/5x + 3/5
slope = -2/5
slope of perpendicular line is negative reciprocal.
m = 5/2
y = mx + b
-3 = (5/2)(2) + b
-3 = 5 + b
b = -8
line =
y = (5/2)x - 8
well, from the previous posting, we already know the slope of 2x+5y=3, is -2/5.
well, perpendicular lines have negative reciprocal slopes,
[tex] \bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}{\stackrel{slope}{-\cfrac{2}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{5}{2}}\qquad \stackrel{negative~reciprocal}{+\cfrac{5}{2}}}\implies \cfrac{5}{2} [/tex]
so we're really looking for an equation of a line whose slope is 5/2 and runs though the same point as before,
[tex] \bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad \qquad \qquad slope = m\implies \cfrac{5}{2}\\\\\\\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-3)=\cfrac{5}{2}(x-2)\implies y+3=\cfrac{5}{2}x-5\\\\\\y=\cfrac{5}{2}x-5-3\implies y=\cfrac{5}{2}-8 [/tex]
