Question

Question 20
If the points A(1,2), B(0,0) and C(a,b) are collinear, then
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = –b

Answer 1

(C)

$\text{Let the given points are}A=(x_1,y_1)=(1,2)B=(x_2,y_2)=(0,0)andC=(x_3,y_3)=(a,b).\because Areaof\Delta ABC\Delta =\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\therefore \Delta =\frac{1}{2}\left[1\right(0-b)+0(b-2)+a(2-0\left)\right]=\frac{1}{2}(-b+0+2a)=\frac{1}{2}(2a-b)\text{Since, the points A(1,2), B(0,0) and C(a,b)}\text{are collinear, then area of triangle should be equal to zero.}\text{i.e, area of}\Delta ABC=0\Rightarrow \frac{1}{2}(2a-b)=0\Rightarrow 2a-b=0\Rightarrow 2a=b\text{Hence, the required relation is 2a = b.}$