SOLUTION: 1.(a)If (p+1)x+y=3 and 3y-(p-1)x=4 are perpendicular to each other,find the value of p. (b)If y+(2p+1)x+3=0 and 8y-(2p-1)x=5 are mutually perpendicular,find the value of p. 2.The

Algebra ->  Coordinate-system -> SOLUTION: 1.(a)If (p+1)x+y=3 and 3y-(p-1)x=4 are perpendicular to each other,find the value of p. (b)If y+(2p+1)x+3=0 and 8y-(2p-1)x=5 are mutually perpendicular,find the value of p. 2.The      Log On


   

Question 550586: 1.(a)If (p+1)x+y=3 and 3y-(p-1)x=4 are perpendicular to each other,find the value of p.
(b)If y+(2p+1)x+3=0 and 8y-(2p-1)x=5 are mutually perpendicular,find the value of p.
2.The co-ordinates of the vertex A of a square ABCD are (1,2) and the equation of the diagonal BD is x+2y=10.Find the equation of the other diagonal and the co-ordinates of the centre of the square.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1.(a)If (p+1)x+y=3 and 3y-(p-1)x=4 are perpendicular to each other, the product of their slopes is -1. I can use that fact to find the value of p.
The slope of %28p%2B1%29x%2By=3 --> y=-%28p%2B1%29x%2B3
is -%28p%2B1%29
The slope of 3y-%28p-1%29x=4 --> 3y=%28p-1%29x%2B4 --> y=%28p-1%29x%2F3%2B4%2F3
is %28p-1%29%2F3
The product of the slopes is
-%28p%2B1%29%28p-1%29%2F3=-1 --> %28p%2B1%29%28p-1%29=3 --> p%5E2-1=3 --> p%5E2=4
There are two solutions p=2 and p=-2
(b)If y+(2p+1)x+3=0 and 8y-(2p-1)x=5 are mutually perpendicular, the product of their slopes is -1. I can use that fact to find the value of p.
The slope of y%2B%282p%2B1%29x%2B3=0 --> y=-%282p%2B1%29x-3
is -%282p%2B1%29
The slope of 8y-%282p-1%29x=5 --> 8y=%282p-1%29x%2B5 --> y=%282p-1%29x%2F8%2B5%2F8
is %282p-1%29%2F8
The product of the slopes is
-%282p%2B1%29%282p-1%29%2F8=-1 --> %282p%2B1%29%282p-1%29=8 --> 4p%5E2-1=8 --> 4p%5E2=9 --> p%5E2=9%2F4
There are two solutions p=3%2F2 and p=-3%2F2
2.The co-ordinates of the vertex A of a square ABCD are (1,2) and the equation of the diagonal BD is x+2y=10.Find the equation of the other diagonal and the co-ordinates of the centre of the square.
I solved this problem recently. It was submitted as question # 550393.