Question 515964
can you please help me find, the equation of the line: passing through (6,6) and perpendicular to 5x-3y-8=0
:
Put the equation in the slope/intercept form to find the slope
5x - 3y - 8 = 0
-3y = -5x + 8
y has to be positive, multiply both sides by -1
3y = 5x - 8
divide both sides by 3
y = {{{5/3}}}x - {{{8/3}}}
:
slope m1 = {{{5/3}}}
the relationship of perpendicular lines: is m1*m2 = -1, to find m2
{{{5/3}}}*m2 = -1
multiply both sides by 3/5
m2 = {{{-3/5}}} is the slope of the perpendicular line
:
Use the point/slope form to find the equation of the perpendicular line
y - y1 = m(x - x1)
x1 = 6, y1 = 6, m=-3/5
:
y - 6 = {{{-3/5}}}(x - 6)
y - 6 = {{{-3/5}}}x - {{{-18/5}}}
y - 6 = {{{-3/5}}}x + 3.6
y  = {{{-3/5}}}x + 3.6 + 6
y = {{{-3/5}}}x + 9.6; the equation of the perpendicular line
:
Check this, substitute 6 for x in the equation and ensure that y = 6