Question 565612
write an equation that has an x intercept of 3 and is perpendicular to the graph of 5x-3y=2
:
Put the equation in the slope/intercept form to find the slope (y = mx + b)
5x - 3y = 2
- 3y = -5x + 2
Multiply both sides by -1
3y = 5x - 2
divide both sides by 3
y = {{{5/3}}}x - {{{2/3}}}
Slope m1 = {{{5/3}}}
:
The relationship of slope in perpendicular lines is: m1*m2 = -1
Find m2
{{{5/3}}}*m2 = -1
m2 = {{{-3/5}}} is the slope of the perpendicular line
:
y = {{{-3/5}}}x + b
The x intercept occurs when y = 0, find b
{{{-3/5}}}*3 + b = 0
{{{-9/5}}} + b = 0
b = {{{9/5}}}
therefore
y - {{{-3/5}}}x + {{{9/5}}} is the equation of the perpendicular line with an x intercept of 3.
:
Graphically
{{{ graph( 300, 300, -4, 4, -4, 4, (5/3)x-(2/3), (-3/5)x+1.8) }}}