Class 10 Ex 7.2

Co-ordinate Geometry Class 10

Question 1 from Class 10 Ex. 7.2

Find the co-ordinate of the point which divides (-1,7) and (4,-3) in the ratio 2:3.

Solution: Co-ordinate of the point which divides the line joing the points (-1,7) and (4,-3) in the ratio 2:3 is

= ( \cfrac{4.2 + (-1).3}{2 + 3} , \cfrac{(-3).2 + 7.3}{2 + 3} )

= ( \cfrac{5}{5} , \cfrac{15}{5} )

= ( 1, 3 )

So, The required point is ( 1, 3 ).

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Question 2 from Class 10 Ex 7.2

Find the co-ordinate of the points of trisection of the line segment joining ( 4, -1 ) and (-2, -3).

Solution: Let P and Q be the points the points of trisection of the line segment joining A( 4, -1 ) and B(-2, -3)

For P, ratio is AP : PB= 1:2

So, co-ordinate of P is

         =( \cfrac{(-2).1 + 4.2}{1 + 2} , \cfrac{(-3).1 + (-1).2}{1 + 2} )

        = ( \cfrac{6}{3} , \cfrac{-5}{3} )

        = ( 2 , – \cfrac{5}{3} )

For Q point ratio is, AQ : QB = 2 : 1

So, co-ordinate of Q is

          = ( \cfrac{(-2).2 + 4.1}{2 + 1} , \cfrac{(-3).2 + (-1).1}{2 + 1} )

         = ( \cfrac{0}{3} , \cfrac{-7}{3} )

         = ( 0, – \cfrac{7}{3} )

Question 3 from class 10 Ex 7.2

 Find the ratio in which the line segment joining the points ( -3, 10 ) and (6,-8) is divided by (-1,6)

Solution: Let the ratio be m:n

So, Co-ordinate of the point which intersect (-3,10) and (6,-8) in m:n ratio is = (-1,6)

⇒ ( \cfrac{6m + (-3)n}{m+n} , \cfrac{(-8)m + 10n}{m+n} ) = ( -1, 6 )

Therefore, \cfrac{6m -3n}{m+n} = -1

⇒ 6m – 3n = -m -n

⇒ 7m = 2n OR \cfrac{m}{n} = \cfrac{2}{7}

So, m : n = 2 : 7

Hence, Required ratio is 2 : 7

Question 5 from Class 10 Ex 7.2 

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 co-ordinate of the point of division.

Solution: Let the point P( x, 0 ) which is on x-axis, divides the line joining A( 1, -5 ) and B( -4, 5 ) in m:n ratio.

So, ( \cfrac{-4.m + 1.n}{m+n} , \cfrac{5.m + (-5).n}{m+n} ) = ( x, 0 )

Hence, \cfrac{-4.m + 1.n}{m+n} = x ————(i)

and \cfrac{5.m -5.n}{m+n} = 0

⇒ 5m – 5n = 0

⇒ 5m = 5n

Hence m : n = 5 : 5 = 1 : 1

Now, From equation (i), by putting m = 1 & n = 1 we get,

\cfrac{-4.1 + 1.1}{1+1} = x

⇒ x = \cfrac{-3}{2}

Hence, Required co-ordinate of the point on the x axis is (0, \cfrac{-3}{2} ) and this point divides the given line in 1 : 1 ratio.

Question 6 from class 10 Ex 7.2

 If ( 1, 2 ), ( 4, y ), ( x, 6 ) and ( 3, 5 ) are the vertices of a parallelogram taken in order, find x and y.

Solution:

The given points ( 1, 2 ), ( 4, y ), ( x, 6 ) and ( 3, 5 ) are the vertices of a parallelogram taken in order.

We know that, the diagonals of a parallelogram bisects each other. Which means, mid points of the diagonals are same.

So,

Mid point of the line joining (1,2) and (x,6) = Mid point of the line joining (4,y) and (3,5)

⇒( \cfrac{1+x}{2} , \cfrac{2+6}{2} ) = ( \cfrac{4+3}{2} , \cfrac{y+5}{2} )

Hence, \cfrac{1+x}{2}[/latex ] = [latex]\cfrac{7}{2}

                   ⇒ 2 + 2x = 14

                   ⇒ 2x = 12

                   ⇒ x = 6

And, \cfrac{2+6}{2} = \cfrac{y+5}{2}

                 ⇒2y + 10 = 16

                 ⇒ 2y = 6

                 ⇒ y = 3

Hence, value of x = 6 and y = 3

Question 7 from class 10 Ex 7.2

 Find the Co-ordinate of the point A where AB is the diameter of a circle whose centre is (2, -3) and B is (1,4).

Solution:

Let, (x,y) be the Co-ordinate of the point A.

We know, centre is the mid point of diameter.

So, Mid point of (x,y) and (1,4) = (2,-3)

⇒ ( \cfrac{x + 1}{2} , \cfrac{y + 4}{2} = ( 2, -3)

So, \cfrac{x + 1}{2} = 2

                    ⇒ x + 1 = 4  ⇒ x = 3

And, \cfrac{y + 4}{2} = -3

                ⇒ y + 4 = -6

                ⇒ y = -10

Hence, required value of x = 3 & y = -10

Question 8 from Class 10 Ex 7.2

 If A and B are (-2, -2) and (2, -4) respectively, find the Co-ordinate of P such that AP = \cfrac{3}{7}AB and P lies on the line segment AB

Solution: Here,

AP = \cfrac{3}{7}AB

Clearly, PB = \cfrac{4}{7}AB

Hence, AP : PB = 3 : 4

So, Co-ordinate of the point which bisects A( -2, -2 ) and B( 2, -4 ) at 3 : 4 is

  = ( \cfrac{2.3 + (-2).4}{3 + 4} , \cfrac{(-4).3 + (-2).4}{3 + 4} )

         = ( \cfrac{-2}{7} , \cfrac{-20}{7} )

Hence, required Co-ordinate of the point A( \cfrac{-2}{7} , \cfrac{-20}{7} )

Question 9 from Class 10 Ex 7.2

 Find the Co-ordinate of the points which divides the line segment joining A( -2, 2 ) and B( 2, 8 ) into four equal parts.

Solution:

Let the points be P, Q, R be the points which intersects the line AB into four parts.

For point P, AP: PB = 1 : 3

So, Co-ordinate of P is

= ( \cfrac{2.1 + (-2).3}{1 + 3} , \cfrac{8.1 + 2.3}{1 + 3} )

          = ( \cfrac{-4}{4} , \cfrac{14}{4} )

          = ( -1 , \cfrac{7}{2} )

For point Q, AQ : QB = 2 : 2

So, Co-ordinate of Q is

        = ( \cfrac{2.2 + (-2).2}{2 + 2} , \cfrac{8.2 + 2.2}{2 + 2} )

       = ( \cfrac{0}{4} , \cfrac{20}{4} )

       = ( 0 , 5 )

For Point R, AR : RB = 3 : 1

So, Co-ordinate of R is

= ( \cfrac{2.3 + (-2).1}{3 + 1} , \cfrac{8.3 + 2.1}{3 + 1} )

        = ( \cfrac{4}{4} , \cfrac{26}{4} )

       = ( 1 , \cfrac{13}{2} )

Therefore, The required points are ( -1 , \cfrac{7}{2} ) , ( 0, 5 ) and ( 1 , \cfrac{13}{2} ) respectively.

Question 10 from class 10 Ex 7.2

Find the area of a rhombus if its vertices are  ( 3, 0), ( 4, 5), ( -1, 4) and ( -2, -1) taken in order.

Solution: Let the four points be A( 3, 0), B( 4, 5), C( -1, 4) and D( -2, -1)

Hence, AC = \sqrt{ (3 - (-1))^2 + (0 - 4)^2}

                   = \sqrt{ 4^2 + 4^2}

                   = 4 \sqrt{2}

            BD = \sqrt{ (4 - (-2))^2 + (5 - (-1))^2}

                   = \sqrt{ 6^2 + 6^2}

                   = 6 \sqrt{2}

Therefore, Area of the Rhombus is

         = \cfrac{1}{2} × AC × BD

         = \cfrac{1}{2} × 4 \sqrt{2} × 6 \sqrt{2}

         = 24 sq. units

Hence, Required area of the Rhombus is 24 Sq. units.