Question 154830
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+4y-2 = 0}}} Subtract px from both sides of the equation.
{{{4y-2 = -px}}} Add 2 to both sides.
{{{4y = -px+2}}} Finally, divide both sides by 4.
{{{y = (-p/4)x+2/4}}} Simplify.
{{{y = (-p/4)x+1/2}}} Here, the slope is {{{m = (-p/4)}}}
Second equation:
{{{2x-y+p = 0}}} Add y to both sides.
{{{2x+p = y}}} or
{{{y = 2x+p}}} Here, the slope is {{{m = 2}}}
The negative reciprocal of 2 is {{{-1/2)}}} and this should equal the slope of the first equation, {{{(-p/4)}}}, so...
{{{(-p/4) = -1/2}}} Simplify and solve for p. Multiply both sides by -4.
{{{p = 4/2}}} Simplify.
{{{p = 2}}}
Check, are the slopes negative reciprocals?
{{{m = (-p/4)}}} Substitute p = 2.
{{{m = (-2/4)}}} Simplify.
{{{m = -1/2}}} ...and this is the negative reciprocal of 2.