Question

1.test whether the points (-3,4),(2,-5),(11,18) are collinear or not
2.if p (x,y) is a point on the line joining the points A(a,0) and B(0,b), show that x/a +y/b=1
3.if A,B,C are the points (0,5),(3,1) and 5,7 respectively and D,E, and F are the midpoints of BC,CA ,AB respectively.show that the area of triangle ABC is four times that of triangle DEF
4. Find the area of the triangle formed by the points (p+1,1),(2p+1,3) and (2p+2,2p) and show that these points are collinear if p=2 or -1/2
5.for what values of k are the points (k,2-2k),(-k+1,2k) and (-4-k,6-2k) are collinear

Answer 1

1) NO
slope of AB = -9/5    slope of BC = 23/9  slope of CA = 14/14  

2) slope of AB = -b/a    so  (y-b)/x = -b/a
            ay -ab = -bx      so answer   x/a+y/b = 1
 
3)   AB = √(4²+3²) = 5       equation of AB =    y-5 = (-4/3) x
                      4x +3 y - 15  =0 
          Altitude of ABC from C to AB :   (20+21-15) / (√3²+4² ) = 26/5 = 5.2
        area = 5 * 5.2 /2 = 13

      D=(4,4)  E = (2.5,  6)  F = (1.5, 3)
          DE = 2.5       EF = √10   FD = √7.25 
      equation of EF = y-3 = 3 (x-1.5)      => y - 3 x +1.5 =0
            Altitude from D on to EF =  | 4 - 12 + 1.5 | /√10 = 6.5/√10
            Area of DEF =  √10 *  6.5/√10 /2 = 3.25

           4 times area of DEF = 13

4)
        equation of AB =      p( y -1) = (x-p-1) 2
                       p y -2 x +p +2 = 0
         Altitude of C onto AB  =    | 2p² - 4p - 4 +p + 2 | / √(p²+4)
                            = | 2 p² - 3 p - 2 | /  √(p²+4)

          Altitude is 0, if p= 2 or -1/2.  so area is 0.

5)
 you can do the same way as i did in the 4th above.