Match the items of column-1 with items of column-2 and choose the correct option.

(A)	If $(8,1),(k,-4),(2,-5)$ are collinear, then k=	(1)	4
(B)	The area of the triangle formed by $(2,3),(-1,0)and(2,-4)$  in sq. units is	(2)	7
(C)	The area of $(t,t-2),(t+2,t+2),(t+3,t)$ is in sq. units is	(3)	3
(D)	If the area of $\Delta ABC=21$ sq. units, the area of $\Delta AGB$ ($G$ is centroid) is	(4)	10.5

(A) $A(8,1),B(k,-4),C(2,-5)$ are collinear

Slope of $AB=$ Slope of $BC$

$\frac{y_2-y_1}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}$

$\frac{-4-1}{k-8}=\frac{-5+4}{2-k}$

$\frac{-5}{k-8}=\frac{-1}{2-k}$

$10-5k=1k-8$

$10+8=1k+5k$

$6k=18$

$k=\frac{18}{6}$

$k=3$

(B) Given $A(2,3),B(-1,0),C(2,-4)$

Area $\Delta ABC=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\frac{1}{2}|2(0+4)+(-1)(-4-3)+2(3-0)|$

$=\frac{1}{2}|8+7+6|$

$=\frac{1}{2}\left|21\right|$

$=\frac{21}{2}=10.5sq.cm.$

(C) $(t,t-2),(t+2,t+2),(t+3,t)$

Area $=\frac{1}{2}|t(t+2-t)+(t+2)(t-t+2)+(t+3)(t-2-t-2)|$

$=\frac{1}{2}|2t+2t+4-4t-12|$

$=\frac{1}{2}|-8|=4$.

(D)

$\Delta ABC=21$ sq. units

$ar\Delta AGB=\frac{1}{3}\left[ar\Delta ABC\right]$

$=\frac{1}{3}\times 21$

$=7$ units.