Question

Find an equation in slope intercept form for the line through (2,-3) and perpendicular to the line 2x+5y=3

Answer 1

well, from the previous posting, we already know the slope of 2x+5y=3, is -2/5.

well, perpendicular lines have negative reciprocal slopes,

`\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}`

so we're really looking for an equation of a line whose slope is 5/2 and runs though the same point as before,

`\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`

Answer 2

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