Question

if the points (3,-1),(p,0),(1,-2) are collinear ,then find the value of 'p'.​

Answer 1

Step-by-step explanation:

Again on squaring both the sides

(p2– 6p + 18) (p2+ 2p + 10) = (–p2+ 2p + 12)2

p4 + 2p3 +10p2 – 6p3–12 p2 – 60p +18p2 + 36p +180 = p4 + 4p2 + 144 – 4p3 + 48p – 24p2

p4 – 4p3 +16p2 – 24p +180 = p4 – 4p3 – 20p2 + 48p + 144

36p2 – 72p + 36 = 0

p2 – 2p +1 = 0

(P – 1)2 = 0

p– 1 = 0

p = 1

Answer 2

Answer:

p=5

Step-by-step explanation:

$ar \: .triangle \: abc = \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) \\ 0 = \frac{1}{2} (3(0 - ( - 2) )+ p( - 2 - ( - 1)) + 1( - 1 - 0))) \\ 0 = 1(6 - p - 1) \\ 0 = 5 - p \\ - 5 = - p \\ p = 5$