SOLUTION: The lines px+4y-2=0 and 2x-y+p=0 are perpendicular. Find the value of p.

Algebra ->  Linear-equations -> SOLUTION: The lines px+4y-2=0 and 2x-y+p=0 are perpendicular. Find the value of p.      Log On


   

Question 154830: The lines px+4y-2=0 and 2x-y+p=0 are perpendicular. Find the value of p.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
If the lines are perpendicular, then their slopes are negative reciprocals.
Let's put the two equations into the slope-intercept form: y = mx+b
First equation:
px%2B4y-2+=+0 Subtract px from both sides of the equation.
4y-2+=+-px Add 2 to both sides.
4y+=+-px%2B2 Finally, divide both sides by 4.
y+=+%28-p%2F4%29x%2B2%2F4 Simplify.
y+=+%28-p%2F4%29x%2B1%2F2 Here, the slope is m+=+%28-p%2F4%29
Second equation:
2x-y%2Bp+=+0 Add y to both sides.
2x%2Bp+=+y or
y+=+2x%2Bp Here, the slope is m+=+2
The negative reciprocal of 2 is -1%2F2%29 and this should equal the slope of the first equation, %28-p%2F4%29, so...
%28-p%2F4%29+=+-1%2F2 Simplify and solve for p. Multiply both sides by -4.
p+=+4%2F2 Simplify.
p+=+2
Check, are the slopes negative reciprocals?
m+=+%28-p%2F4%29 Substitute p = 2.
m+=+%28-2%2F4%29 Simplify.
m+=+-1%2F2 ...and this is the negative reciprocal of 2.