I need step-by-step answers so I can fully understand.
What is the slope of a line perpendicular to 3x-2y+4?
Tutors, please sign in to answer this question.
First get the equation in slope-intercept form (y=mx+b, where m=slope):
3x - 2y + 4 = 0
-2y +4 = -3x
-2y = -3x - 4
-2 -2 -2
y = (3/2)x - (-2)
y = (3/2)x + 2
The slope of the line for this equation is the number multiplied to the x. So, m (slope) is 3/2.
Now, a perpendicular slope will be the negative reciprocal. The negative recipriocal of 3/2 = -2/3.
Assuming the equation is 3x-2y+4=0 then the equation of the line, solving for y in terms of x, gives y=3/2x+2 the slope of this line is 3/2 then the slope of a line that is perpendicular is -2/3 .
An easy way to see this is to draw the line. By definition the slope, m=tan(α) where α is the angle the line makes with the x axis. Let L=slope of the perpendicular line then L=tan(β) but the two lines with the x axis form a right triangle hence α+π-β=π/2 so β=π/2+α and L=tan(β)=tan(π/2+α)=-cot(α) = -1/tan(α)=-1/m.
In our problem m=3/2 so L=-2/3.
The geometry becomes clear if you make a sketch of the two lines. Hope this helps