Question 1025651
The shortcut to proving it is to know what to recognize in the standard form equation for a line.  The slopes of your given lines in order are  {{{1/p}}} and {{{1/q}}}.  The lines to be perpendicular will require  {{{(1/p)(1/p)=-1}}}.  Multiply left and right members by {{{pq}}}.  
.... there, you have it.



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So you want a little more information for this.
Basic fact is,  if product of the slopes of two lines is negative ONE, then the two lines are perpendicular.


Put each equation into slope-intercept form.
{{{x-py+ap^2=0}}}
{{{-py=-x-ap^2}}}
{{{y=(1/p)x+ap^2}}}
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{{{x-qy+aq^2=0}}}
{{{-qy=-x-aq^2}}}
{{{y=(1/q)x+aq^2}}}
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Note the slopes of the two lines are {{{1/p}}} and {{{1/q}}}.
The two lines are GIVEN as perpendicular, and this means according to the basic fact about perpendicular lines in the plane,  {{{(1/p)(1/q)=-1}}}.


{{{(1/p)(1/q)*pq=-1*(pq)}}}

{{{1=-1*pq}}}


{{{1*(-1)=-1*pq*(-1)}}}


{{{-1=pq}}}-------done.