Question

1)find the value of p for which the points (-5,1),(1,p) and (4,-2) are collinear.
2)prove that the points A(-3,3),B(4,3),C(4,-2) and D(-3,-2)are the vertices of the rectangle.
3)find the point on x-axis which is equidistant from points (-1,0) and (5,0)
guyzzz!!!!!step wise answers plzzz.....need it urgently <hope u understand>

Answer 1

1. slope of AB = (p-1)/6 = slope of AC = -3/9 = -1/3
     p-1 = -2   => p = -1

2.  AB = 7    BC = 5  CD = 7  DA = 5
      Slope AB = 0  = parallel to x axis  slope  BC = infinity = parallel to y axis
       so perpendicular.  hence proved

3.  let the point be  (x,0).    then   (x+1)^2 = (5-x )^2
                  so  2x = 4  ie.  x = 2      

Answer 2

1.
The distance between the (-5,1) & (1,P) is equal to the distance between (1,P) & (4,-2)
therefore,
$\sqrt{(-5-1)^2+(1-P)^2} = \sqrt{(1-4)^2 + (p-(-2))^2}$
$\sqrt{-6^2+(1-p)^2} = \sqrt{-3^2 + (p+2)^2}$
Both the square root cancels
36 + 1 + P^2 - 2P = 9 + P^2 +4 +4P
Both the P^2 cancels
37 - 2P = 13 +4P
6P = 37-13
P=24
P=24/6
P=4.