1.   Find the area of the triangle whose vertices are :
(i)   (2, 3), (–1, 0), (2, – 4)


area of ∆ ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
              = ${1/2}{[2 (0 - (-4)) + (-1) (-4 - 3) + 2 (3 – 0)]}$
              = ${1/2}{[2 (4) + (-1) (-7) + 2 (3)]}$
              = ${1/2}{[8 + 7 + 6]}$
              = ${21/2}$  sq. units


(ii)  (–5, –1), (3, –5), (5, 2)


area of ∆ ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
              = ${1/2}{[-5 (-5 - 2) + 3 (2 - (-1)) + 5 (-1 – (-5))]}$
              = ${1/2}{[-5 (-7) + 3 (3) + 5 (4)]}$
              = ${1/2}{[35 + 9 + 20]}$
              = ${64/2}$
              = $32$ sq. units

2.   In each of the following find the value of ‘k’, for which the points are collinear. 
(i)   (7, –2), (5, 1), (3, k)


Since the points are collinear, so the area of ∆ is zero.
area of ∆ ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
           0  = ${1/2}{[7 (1 - k) + 5 (k - (-2)) + 3 (-2 – 1)]}$
           0  = ${[(7 - 7k) + (5k +10) -9]}$
           0  = ${[8-2k]}$
           $2k=8$
           $k=4$


(ii)  (8, 1), (k, – 4), (2, –5)


Since the points are collinear, so the area of ∆ is zero.
area of ∆ ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
            0= ${1/2}{[8 (-4 - (-5)) + k (-5 - 1) + 2 (1 – (-4))]}$ 
            0= ${1/2}{[8 (1) + k (-6) + 2 (5)]}$ 
            0= ${[8 -6k +10]}$ 
            $6k=18$
            $k=3$

3.   Find the area of the triangle formed by joining the mid-points of the sides of the triangle
 whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
 









A(0,-1)
B(2,1)
C(0,3)
P(1,0)
Q(1,2)
R(0,1)

 
 Since P, Q and R are the mid points of the side AB,BC and CA of ∆ ABC,
 therefore the coordinates of P,Q and R are ,
 P = $((0+2)/2 , (-1+1)/2) $ =$(1,0)$
 Q = $((2+0)/2 , (1+3)/2) $ = $(1,2)$
 R = $((0+0)/2 , (3-1)/2) $ = $(0,1)$

area of ∆ ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
			=${1/2}{[0 (1 - 3) + 2 (3 - (-1)) + 0(-1 – 1)]}$ 
			= $1/2(8)$
			= $4$ sq. units

area of ∆PQR = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$	
             =${1/2}{[1 (2 - 1) + 1 (1 - 0) + 0 (0 – 2)]}$		   
             =${1/2}{[1+1+0]}$
			 = $1$  sq. units
			 
Ratio ∆ ABC:∆PQR = 1:4


4.   Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).








A(-4,-2)
B(-3,-5)
C(3,-2)
D(2,3)


 
Area of quadrilateral ABCD= Area of ∆ABC +Area of ∆ACD

Area of ∆ABC = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$
             =${1/2}{[-4 (-5 - (-2)) + (-3) (-2 - (-2)) + 3 (-2 – (-5))]}$
             =${1/2}{[-4 (-3) + (-3) (0) + 3 (3)]}$
             =${1/2}{[12+9]}$
             =${21/2}$

Area of ∆ACD = ${1/2}{[-4 (-2 - 3) + 3 (3 - (-2)) + 2 (-2 – (-2))]}$
             =${1/2}{[-4 (-5) + 3 (5) + 2 (0)]}$
             =${1/2}{[20+15+0]}$
             =${35/2}$
			 
Area of quadrilateral ABCD= $21/2$ +$35/2$
                          = $56/2$
						  =$28$ sq. unit
			 

5.   A median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ ABC whose 
     vertices are A(4, – 6), B(3, –2) and C(5, 2).








A(4,-6)
B(3,-2)
C(5,2)
D(4,0)


 

In the ∆ABC , AD is the median . So the coordinatres of D are,
 D= $({3+5}/2,{-2+2}/2)$ = $(4,0)$

Area of ∆ABC = ${1/2}{[4 (-2 - 2) + 3 (2 - (-6)) + 5 (-6 – (-2))]}$
             = ${1/2}{[4 (-4) + 3 (8) + 5 (-4)]}$
             = ${1/2}{[-16 + 24 -20]}$
             = ${1/2}{[-36+24]}$
             = ${1/2}{[-12]}$
             = ${6}$ sq. unit
              
Area of ∆ABD = ${1/2}{[4 (-2 -0) + 3 (0 - (-6)) + 4 (-6 – (-2))]}$
             =${1/2}{[4 (-2 ) + 3 (6) + 4 (-4)]}$
             =${1/2}{[-8 + 18 -16]}$
             =${1/2}{[-24 + 18]}$
             =${1/2}{[-6]}$
             =${3}$ sq. unit
			 
			 
Area of ∆ACD = ${1/2}{[4 (2 - 0) + 5(0 - (-6)) + 4 (-6 – 2)]}$ 
             = ${1/2}{[4 (2) + 5(6) + 4 (-8)]}$ 
             = ${1/2}{[8 + 30 -32]}$ 
             = ${1/2}{[6]}$ 
             = ${3}$ sq. unit.
			 
Since ,
Area of ∆ABC = Area of ∆ABD + Area of ∆ACD
Therfore median of a triangle divides it into two triangles of equal areas