5x + 3y = - 53
the equation of a line in ' slope- intercept form ' is y = mx + c
where m is the slope and c the y-intercept
rearrange 3x - 5y = - 15 into this form to obtain m → (subtract 3x from both sides)
- 5y = - 3x - 15 → divide all terms by - 5 )
y = [tex]\frac{3}{5}[/tex] x + 3 → in slope-intercept form with m = [tex]\frac{3}{5}[/tex]
given a line with slope m then the slope m₁ of a line perpendicular to it is
m₁ = - [tex]\frac{1}{m}[/tex] = - 1 ÷ [tex]\frac{3}{5}[/tex] = - [tex]\frac{5}{3}[/tex]
partial equation is y = - [tex]\frac{5}{3}[/tex] x + c
to find c substitute ( - 7, - 6) into the partial equation
- 6 = [tex]\frac{35}{3}[/tex] + c ⇒ c = - 6 - [tex]\frac{35}{3}[/tex] = - [tex]\frac{53}{3}[/tex]
y = - [tex]\frac{5}{3}[/tex] x - [tex]\frac{53}{3}[/tex] → in slope intercept form
multiply all terms by 3
3y = - 5x - 53 → ( add 5x to both sides )
5x + 3y = - 53 → in standard form
