Question

If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then

Answer 1

Answer:

b is twice of a

Step-by-step explanation:

area =1/2 {(1×0)+(0×b)+(a×2)} - { (2×0)+(0×a)+(b×1)}

0= {0+0+2a} - {0+0+b}

0= 2a-b

2a=b

Answer 2

$\bf\huge\underline{Question}$

If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then

$\bf\huge\underline{Answer}$

Let the given points are A(x1, y1) = (1, 2), O(x2, y2) = (0, 0) and C(x3, y3) = (a, b).

Therefore, Area of ∆AOC

=> $\dfrac{1}{2}$ [x1(y2 - y3) + x2(y3 - y1)+x3(y1+y2)]

=> $\dfrac{1}{2}$ [1(0 - b) + 0(b - 2) + a(2 - 0)]

=> $\dfrac{1}{2}$ [(-b + 0 + 2a) = $\dfrac{1}{2}$ (2a - b)

Since, the points A(1, 2), O(0, 0) and C(a, b) are collinear, then area of ∆AOC should be equal to zero

i.e., area of ∆AOC = 0

=> $\dfrac{1}{2}$ (2a - b) = 0

=> 2a - b = 0

=> 2a = b

Hence, the required relation is 2a = b