<PRE><B>Question 1172152</B>
the equation of AB: slope is 3/2 and point slope y-y1=m(x-x1), m slope (x1, y1) point is 
y+5=(3/2)(x+1) 
y=(3/2)x-7/2
The line perpendicular to AB has slope -2/3.
AP is 4 x units or x=+3 from A and 6 y units or +1 from A.
so P is at (3, 1)
So AP has equation y-1=(-2/3) (x-3)
y=(-2/3)x+3
graph these
{{{graph(500,500,-15,15,-15,15,11,1,(1.5x-7/2),(-2/3)x+3)}}}

Point Q has y=11 and is on line y=(-2/3)x+3, so 8=(-2/3)x and x=-12
(-12, 11)
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