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Coordinate Geometry : Exercise - 7.2 (Mathematics NCERT Class 10th)


Q.1     Find the coordinates of the point which divides join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Sol.       Let P(x, y) be the required point.
             By Section formula 
             P\left( {x,y} \right) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]
             Where, m = 2 and n = 3
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             Then, x = {{2 \times 4 + 3 \times - 1} \over {2 + 3}}
             and, y = {{2 \times - 3 + 3 \times 7} \over {2 + 3}}
             \Rightarrow x = {{8 - 3} \over 5}
             and  y = {{ - 6 + 21} \over 5}
              \Rightarrow x = 1 and y = 3
             So, the coordinates of P are (1, 3).


Q.2     Find the coordinates of the points of trisection of the line segment joining (4,–1) and (–2,–3).
Sol.       Let A(4,–1) and B(–2,–3) be the points of trisection of P and Q.
             Then, AP = PQ= QB = k (say).
             Therefore, PB = PQ + QB = 2k
             and, AQ = AP + PQ = 2k
              \Rightarrow AP : PB = k : 2k = 1 : 2
             and AQ : QB = 2k : k = 2 : 1

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             So, P divides AB internally in the ratio 1 : 2, while Q divides AB internally in the ratio 2 : 1.
             Thus, the coordinates of
            P are
 \left( {{{1 \times - 2 + 2 \times 4} \over {1 + 2}},{{1 \times - 3 + 2 \times - 1} \over {1 + 2}}} \right) = \left( {{{ - 2 + 8} \over 3},{{ - 3 - 2} \over 3}} \right) = \left( {2,{{ - 5} \over 3}} \right)
             and, the coordinates of Q are  \left( {{{2 \times - 2 + 1 \times 4} \over {2 + 1}},{{2 \times - 3 + 1 \times - 1} \over {2 + 1}}} \right) = \left( {{{ - 4 + 4} \over 3},{{ - 6 - 1} \over 3}} \right) = \left( {0,{{ - 7} \over 3}} \right)
             Hence, the two points of trisection are \left( {2,{{ - 5} \over 3}} \right) and \left( {0,{{ - 7} \over 3}} \right).


Q.3     To conduct Sports Day activities in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure. Niharika runs {1 \over 4} th the distance AD on the 2nd line and posts a green flag. Preet runs {1 \over 5}th  the distance AD on the eighth line and posts a red flag. What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line (segment) joining the two flags, where should she post her flag?
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Sol.     Clearly from the figure, the position of green flag posted by Niharika is given by P\left( {2,{1 \over 4} \times 100}\right),i.e., P(2, 25)
            and that of red flag posted by Preet is given by Q\left( {8,{1 \over 5} \times 100} \right) 
i.e., Q(8, 20).
            Now, PQ = \sqrt {{{\left( {8 - 2} \right)}^2} + {{\left( {20 - 25} \right)}^2}}
                            = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 5} \right)}^2}}
                            = \sqrt {36 + 25} = \sqrt {61}
           Therefore, the distance between the flags = \sqrt {61} metres
           Let M be the position of the blue flag posted by Rashmi in the halfway of line segment PQ.

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         Therefore, M is the given by \left( {{{2 + 8} \over 2},{{25 + 20} \over 2}} \right) 
or \left( {{{10} \over 2},{{45} \over 2}} \right),i.e., (5, 22.5) .
         T
hus, the blue flag is on the fifth line at a distance 22.5m above it.


Q.4      Find the ratio in which the line segment joining the points of (–3, 10) and (6, – 8) is divided by (–1, 6).
Sol.       Let the point P(–1, 6) divide the line joining A(–3, 10) and B(6, –8) in the ratio k : 1. Then, the
             coordinates of P are 
\left( {{{6k - 3} \over {k + 1}},{{ - 8k + 10} \over {k + 1}}} \right)
             But, the coordinates of P are given as (–1, 6).

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             Therefore, {{{6k - 3} \over {k + 1}} = -1} and {{{ - 8k + 10} \over {k + 1}} = 6}
              \Rightarrow 6k - 3 = - k - 1 and  - 8k + 10 = 6k + 6
              \Rightarrow 6k + k = - 1 + 3 and  - 8k - 6k = 6 - 10
              \Rightarrow 7k = 2 and  - 14k = - 4
              \Rightarrow k = {2 \over 7}
             Hence, the point P divides AB in the ratio 2 : 7.


Q.5      Find the ratio in which the line segment joining A(1, –5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Sol.       Let the required ratio be k : 1. Then, the coordinates of the point P of division are \left( {{{ - 4k + 1} \over {k + 1}},{{ 5k - 5} \over {k + 1}}} \right).
             But it is a point on x-axis on which y - coordinate of every point is zero.

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             Therefore, {{{5k - 5} \over {k + 1}} = 0}
              \Rightarrow 5k - 5 = 0
              \Rightarrow 5k = 5
              \Rightarrow k = 1
            Thus, the required ratio is 1 : 1 and the point of division P is given by
            \left( {{{ - 4 \times 1 + 1} \over {1 + 1}},0} \right),i.e., \left( {{{ - 4 + 1} \over 2},0} \right) i.e., \left( {{{ - 3} \over 2},0} \right)


Q.6      If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find 
x and y.
Sol.        Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
              Since the diagonals of a parallelogram bisect each other,

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              Therefore, {{x + 1} \over 2} = {{3 + 4} \over 2}
               \Rightarrow x + 1 = 7
               \Rightarrow x = 6
              and, {{5 + y} \over 2} = {{6 + 2} \over 2}
               \Rightarrow 5 + y = 8
               \Rightarrow y = 3
              Hence, x = 6 and y = 3


Q.7      Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).
Sol.       Let AB be a diameter of the circle having its centre at C(2, –3) such that the coordinates of end B are (1, 4).

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             Let the coordinates of A be (x, y).
             Since C is the mid - point of AB , therefore
             {{x + 1} \over 2} = 2 \Rightarrow x + 1 = 4 \Rightarrow x = 3
             and, {{y + 4} \over 2} = - 3 \Rightarrow y + 4 = - 6 \Rightarrow y = - 10
             Hence, the coordinates of A are (3, –10) .


Q.8      If A and B are (–2, –2) and (2, – 4) respectively, find the coordinates of P such that AP = {3 \over 7} AB and P lies on the line segment AB.
Sol.       We have, AP = {3 \over 7}AB
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             \Rightarrow {{AB} \over {AP}} = {7 \over 3}
             \Rightarrow {{AP + PB} \over {AP}} = {{3 + 4} \over 3}
             \Rightarrow 1 + {{PB} \over {AP}} = 1 + {4 \over 3} \Rightarrow {{PB} \over {AP}} = {4 \over 3}
             \Rightarrow {{AP} \over {PB}} = {3 \over 4} \Rightarrow AP:PB = 3:4
            Let P(x, y) be the point which divides the join of A(–2, – 2) and B(2, – 4) in the ratio 3 : 4.
            Therefore, x = {{3 \times 2 + 4 \times - 2} \over {3 + 4}} = {{6 - 8} \over 7}  = {{ - 2} \over 7}
            and, y = {{3 \times - 4 + 4 \times - 2} \over {3 + 4}} = {{ - 12 - 8} \over 7} = {{ - 20} \over 7}
            Hence, the coordinates of the point P are \left( { - {2 \over 7},{{ - 20} \over 7}} \right).


Q.9     Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Sol.      Let {P_1},{P_2} and {P_3} be the points that divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.
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           Since {P_2} divides the segment into two equal parts
          Therefore, coordinates of {P_2} (i.e., mid - point) are
          \left( {{{ - 2 + 2} \over 2},{{2 + 8} \over 2}} \right), i.e., (0, 5).
          Now, {P_1} divides the line segment A{P_2} into two equal parts.
          Therefore, coordinates of {P_1}(i.e., mid - point) are
          \left( {{{ - 2 + 0} \over 2},{{2 + 5} \over 2}} \right), i.e., \left( { - 1,{7 \over 2}} \right).
             Again, {P_3} is the mid point of line segment {P_2 B}
         Therefore, coordinates of {P_3} are \left( {{{0 + 2} \over 2},{{5 + 8} \over 2}} \right), i.e., \left( {1,{{13} \over 2}} \right).


Q.10    Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2,–1) taken in order.
Sol.        Let A(3, 0), B(4, 5), C(–1, 4) and D(–2,– 1) be the vertices of the rhombus ABCD.
              Diagonal AC = \sqrt {{{\left( { - 1 - 3} \right)}^2} + {{\left( {4 - 0} \right)}^2}}    = \sqrt {16 + 16} = 4\sqrt 2  and, diagonal
        BD = \sqrt {{{\left( { - 2 - 4} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}}  
 = \sqrt {36 + 36} = 6\sqrt 2
              Area of the rhombus ABCD
              = {1 \over 2} \times (Product of lengths of diagonals )
              = {1 \over 2} \times AC \times BD
              = {1 \over 2} \times 4\sqrt 2 \times 6\sqrt 2 sq. units
             = 24 sq. units

 



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