Question 550586
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 {{{(p+1)x+y=3}}} --> {{{y=-(p+1)x+3}}}
is {{{-(p+1)}}}
The slope of {{{3y-(p-1)x=4}}} --> {{{3y=(p-1)x+4}}} --> {{{y=(p-1)x/3+4/3}}}
is {{{(p-1)/3}}}
The product of the slopes is
{{{-(p+1)(p-1)/3=-1}}} --> {{{(p+1)(p-1)=3}}} --> {{{p^2-1=3}}} --> {{{p^2=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+(2p+1)x+3=0}}} --> {{{y=-(2p+1)x-3}}}
is {{{-(2p+1)}}}
The slope of {{{8y-(2p-1)x=5}}} --> {{{8y=(2p-1)x+5}}} --> {{{y=(2p-1)x/8+5/8}}}
is {{{(2p-1)/8}}}
The product of the slopes is
{{{-(2p+1)(2p-1)/8=-1}}} --> {{{(2p+1)(2p-1)=8}}} --> {{{4p^2-1=8}}} --> {{{4p^2=9}}} --> {{{p^2=9/4}}}
There are two solutions {{{p=3/2}}} and {{{p=-3/2}}}
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.