Coordinate Geometry : Exercise - 7.3 (Mathematics NCERT Class 10th)



Q.1     Find the area of the triangle whose vertices are :
             (i) (2, 3), (–1, 0), (2, – 4)             (ii) (– 5,–1), (3, – 5), (5, 2)
Sol.      (i) LetA = \left( {{x_1},{y_1}} \right) = \left( {2,3} \right),B = \left( {{x_2},{y_2}} \right) = \left( { - 1,0} \right)
             and C = \left( {{x_3},{y_3}} \right) = \left( {2, - 4} \right)
             Area of \Delta ABC
              = {1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]
             = {1 \over 2}\left[ {2\left( {0 + 4} \right) + \left( { - 1} \right)\left( { - 4 - 3} \right) + 2\left( {3 - 0} \right)} \right]
             = {1 \over 2}\left( {8 + 7 + 6} \right) = {{21} \over 2}sq.units

            (ii) Let A = \left( {{x_1},{y_1}} \right) = \left( { - 5, - 1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {3, - 5} \right)
             and C = \left( {{x_3},{y_3}} \right) = \left( {5,\,\,2} \right)

           Area of \Delta ABC
            = {1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]
            = {1 \over 2}\left[ { - 5\left( { - 5 - 2} \right) + 3\left( {2 + 1} \right) + 5\left( { - 1 + 5} \right)} \right]
            = {1 \over 2}\left( {35 + 9 + 20} \right) = {1 \over 2} \times 64
           = 32 sq. units


Q.2     In each of the following find the value of 'k' for which the points are collinear : 
          (i) (7, –2), (5, 1), (3, k)     
(ii) (8, 1), (k, – 4), (2, – 5)
Sol.      (i) Let the given points be A = \left( {{x_1},{y_1}} \right) = \left( {7, - 2} \right),B = \left( {{x_2},{y_2}} \right) = \left( {5,\,\,1} \right) and C = \left( {{x_3},{y_3}} \right) = \left( {3,\,k} \right)
            These points lie on a line if
            Area \left( {\Delta ABC} \right) = 0
             \Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0
            \Rightarrow 7\left( {1 - k} \right) + 5\left( {k + 2} \right) + 3\left( { - 2 - 1} \right) = 0
            \Rightarrow 7 - 7k + 5k + 10 - 9 = 0
            \Rightarrow 8 - 2k = 0
            \Rightarrow 2k = 8
            \Rightarrow k = 4
           Hence, the given points are collinear for k = 4

          (ii) Let the given points be A = \left( {{x_1},{y_1}} \right) = \left( {8,1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {k, - 4} \right) and\,C = \left( {{x_3},{y_3}} \right) = \left( {2,{\mkern 1mu} - 5} \right).
          If the given points are collinear, then
          \Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0
           \Rightarrow 8\left( { - 4 + 5} \right) + k\left( { - 5 - 1} \right) + 2\left( {1 + 4} \right) = 0
           \Rightarrow 8 - 6k + 10 = 0
           \Rightarrow – 6k = – 18
           \Rightarrow k = 3
          Hence, the given points are collinear for k = 3.


Q.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.

Sol.     Let A = \left( {{x_1},{y_1}} \right) = \left( {0,\,\, - 1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {2,\,\,1} \right)
            and C = \left( {{x_3},{y_3}} \right) = \left( {0,\,3} \right) be the vertices of {\Delta ABC}.
            Area \left( {\Delta PQR} \right)
             = {1 \over 2}\left[ {1\left( {0 - 1} \right) + 1\left( {1 - 2} \right) + 0\left( {2 - 0} \right)} \right]
             = {1 \over 2}\left( { - 1 - 1 + 0} \right) = - 1
             = 1 sq. unit (numerically)
     

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           Area \left( {\Delta ABC} \right)
          = {1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]
           = {1 \over 2}\left[ {0\left( {1 - 3} \right) + 2\left( {3 + 1} \right) + 0\left( { - 1 - 1} \right)} \right]
           = {1 \over 2}\left( {0 + 8 + 0} \right) = 4 sq. units
           Let P\left( {{{0 + 2} \over 2},{{3 + 1} \over 2}} \right),i.e., (1, 2), Q\left( {{{2 + 0} \over 2},{{1 - 1} \over 2}} \right), i.e.,  (1, 0) and R\left( {{{0 + 0} \over 2},{{3 - 1} \over 2}} \right), i.e., (0, 1) are the vertices
          of {\Delta PQR}  formed by joining the mid-points of the sides of {\Delta ABC}.

          Ratio of the area \left( {\Delta PQR} \right) to the area \left( {\Delta ABC} \right) = 1 : 4.


Q.4     Find the area of the quadrilateral whose vertices, taken in order, are (– 4, –2), (– 3, –5), (3, –2) and (2, 3).
Sol.     Let A(– 4, –2), B (– 3, –5), C (3, –2) and D (2, 3) be the vertices of the quadrilateral ABCD.
            = Area of \Delta ABC + Area of \Delta ACD
             = {1 \over 2}\left[ { - 4\left( { - 5 + 2} \right) - 3\left( { - 2 + 2} \right) + 3\left( { - 2 + 5} \right)} \right]
             + {1 \over 2}\left[ { - 4\left( { - 2 - 3} \right) + 3\left( {3 + 2} \right) + 2\left( { - 2 + 2} \right)} \right]
             = {1 \over 2}\left( {12 - 0 + 9} \right) + {1 \over 2}\left( {20 + 15 + 0} \right)
             = {1 \over 2}\left( {21 + 35} \right) = {1 \over 2} \times 56 = 28 sq. units


Q.5     You have studied in class IX (Chapter 9 Example 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for \Delta ABC whose vertices are A(4, –6), B(3, –2) and C(5, 2).
Sol.     Since AD is the median of {\Delta ABC}, therefore, D is the mid-point of BC. Coordinates of D are
            \left( {{{3 + 5} \over 2},{{2 + 2} \over 2}} \right),i.e., (4, 0).


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            Area of \Delta ADC
             = {1 \over 2}\left[ {4\left( {0 - 2} \right) + 4\left( {2 + 6} \right) + 5\left( { - 6 - 0} \right)} \right]
             = {1 \over 2}\left( { - 8 + 32 - 30} \right) = {1 \over 2} \times - 6 = - 3
            = 3sq. units (numerically)
             = {1 \over 2}\left[ {4\left( { - 2 - 0} \right) + 3\left( {0 + 6} \right) + 4\left( { - 6 + 2} \right)} \right]
             = {1 \over 2}\left( { - 8 + 18 - 16} \right) = {1 \over 2}\left( { - 6} \right) = - 3
            = 3 sq. units (numerically)
           Clearly, area \left( {\Delta ADC} \right) = area \left( {\Delta ABD} \right)
           Hence, the median of the triangle divides it into two triangles of equal areas.

 



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