Exercise 7.1 Class 10 Maths

Co-Ordinate Geometry

Class 10 Maths Ex 7.1 Question 1

Find the distances between following pairs of points:

(i) (2, 3), (4, 1)               (ii) (-5, 7),  (-1, 3)   

(iii) (a, b), (-a, -b)

Solutions: (i) Distance between (2, 3) and (4, 1)

        is    = \sqrt{(4 - 2)² + (1 - 3)²}

              = \sqrt{(2)² + (-2)²}

              = \sqrt{4 + 4} = √8  =  2√2

Therefore, Required Distance is 2√2 unites.

(ii) Diastance between (-5. 7) and (-1, 3)

     is    = \sqrt{(-1 - (-5))² + (3 - 7)²}

           = \sqrt{(-1 + 5)² + (-4)²}

           = \sqrt{(4)² + (-4)²}

           = \sqrt{16 + 16}   = √32  =  4√2

   Therefore, Required Distance = 4√2 unites.

(iii) Distance between (a, b) and (-a, -b)

     is   = \sqrt{ (-a - a)² + (-b - b)²}

          = \sqrt{ (-2a)² + (-2b)²}

          = \sqrt{ 4(a² + b²)}

         = 2 \sqrt{a² + b²}

Therefore, The required distance is 2 \sqrt{a² + b²}   unites.

Class 10 Maths Ex 7.1 Question 2

Find the distance between (0, 0) and (36, 15).

Solution: The distance between the origin and (36, 15)

is = \sqrt{36² + 15²} = \sqrt{1296 + 225}

    = \sqrt{1521} = 39

  So, Required distance between the given points is 39 unites.

Class 10 Maths Ex 7.1 Question 3

Determine the points (1, 5), (2, 3) and (-2, -11) are collinear.

Solution:

First Method:

         Let the points be A(1, 5), B(2, 3) and C(-2, -11)

Now, Distance AB = \sqrt{(2 - 1)² + (3 - 5)²}

                            = \sqrt{(1)² + (-2)²}

                            = \sqrt{5}

Distance BC         = \sqrt{(-2 - 2)² + (-11 - 3)²}

                            = \sqrt{(-4)² + (-14)²}

                            = \sqrt{16 + 196}

                            = \sqrt{212} = 2√53

Distance AC         = \sqrt{(-2 - 1 )² + (-11 - 5 )²}

                            = \sqrt{(-3)² + (-16)²}

                            = \sqrt{9 + 256} = \sqrt{265}

             Clearly, AB + BC ≠ AC here.

     So, The given three are not collinear.

2nd Method:

Here,

   x1 = 1, y1 = 5, x2 = 2, y2 = 3, x3 = -2, y3 = -11

Now,  x1 (y2 –  y3) + x2 (y3 – y1) + x3 (y1 – y2)

          = 1{3 – (-11)} + 2(-11 – 5) + (-2)(5 – 3)

          = 1 (3 + 11) + 2 × (-16) -2 × 2 = 14-32-4

          = -22 ≠ 0

Therefore, The given three points will create a triangle, means the points are not collinear.

Class 10 Maths Ex 7.1 Question 4

Check whether the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Solution: Let the points be A(5, -2), B(6, 4) and C(7, -2)

Distance AB = \sqrt{(6 - 5)² + (4 - (-2) )²}

                    = \sqrt{(1)² + (6)²}

                    = \sqrt{1 + 36}

                   = \sqrt{37}

Distance BC = \sqrt{(7 - 6)² + (-2 - 4 )²}

                    = \sqrt{(1)² + (-6)²}

                    = \sqrt{1 + 36} = \sqrt{37}

         We get, AB = BC

So, The given three points are the vertices of a isosceles triangle.

Class 10 Maths Ex 7.1 Question 6

Name the type of quadrilateral formed if any, by the following points, and give reasons for your answers.

1. (-1, -2), (1, 0), (-1, 2), (-3, 0)
2. (-3, 5), (3, 1), (0, 3), (-1, -4)
3. (4, 5), (7, 6), (4, 3), (1, 2)

Solutions:

(i) let the points be P(-1, -2), Q(1, 0), R(-1, 2) and S(-3, 0)

Distance PQ = \sqrt{(1 - (-1))² + (0 - (-2) )²}

                   = \sqrt{(1 + 1)² + (0 + 2)²}

                   = \sqrt{4 + 4}

                   = \sqrt{8} = 2 \sqrt{2}

Distance QR = \sqrt{(-1 - 1)² + (2 - 0)²}

                   = \sqrt{(-2)² + (2)²}

                   = \sqrt{4 + 4}

                   = 2 \sqrt{2}

Distance RS  = \sqrt{(-3 - (-1))² + (0 - 2)²}

                   = \sqrt{(-3 + 1)² + (-2)²}

                   = \sqrt{4 + 4}  

                   = 2 \sqrt{2}

Distance PS  = \sqrt{(-3 - (-1))² + (0 - (-2))²}

                   = \sqrt{(-3 + 1)² + (0 + 2)²}

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

So, we found that the lenghts of all the sides are equal.

Diagonal PR = \sqrt{(-1 - (-1))² + (2 - (-2))²}

                  = \sqrt{(-1 + 1)² + (2 + 2)²}

                  = \sqrt{0 + 16} = 4

Diagonal QS = \sqrt{(-3 -1)² + (0 - 0)²}

                   = \sqrt{16 + 0}   = 4

Finally, we also found that the lenghts of diagonal are also same.

This is the property of rhombus. So, The four given points will form a rhombus.

(ii) Let the points be P(-3, 5), Q(3, 1), R(0, 3) and S(-1, -4)

Distance PQ = \sqrt{(3 - (-3))² + (1 - 5)²}

                  = \sqrt{(3 + 3)² + (-4)²}

                  = \sqrt{36 + 16}

                    = \sqrt{52} = 2 \sqrt{13}

Distance QR = \sqrt{(0 - 3)² + (3 - 1)²}

                  = \sqrt{(-3)² + (2)²}

                  = \sqrt{9 + 4} = \sqrt{13}

Distance RS = \sqrt{(-1 - 0)² + (-4 -3)²}

                  = \sqrt{(-1)² + (-7)²}

                  = \sqrt{1 + 49}

                  = \sqrt{50} = 5 \sqrt{2}

Distance PS = \sqrt{(-1 - (-3))² + (-4 -5)²}

                  = \sqrt{(-1 + 3)² + (-9)²}

                  = \sqrt{4 + 81}

                  = \sqrt{85}

       Here, All the lenghts are unequal.

So, The given points will not create any quadrilateral.

(iii) Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)

Distance PQ = \sqrt{(7 - 4)² + (6 - 5)²}

                  = \sqrt{(3)² + (1)²}

                  = \sqrt{9 + 1} = \sqrt{10}

Distance QR = \sqrt{(4 - 7)² + (3 - 6)²}

                  = \sqrt{(-3)² + (-3)²}

                  = \sqrt{9 + 9} = 3 \sqrt{2}

Distance RS = \sqrt{(1 - 4)² + (2 - 3)²}

                  = \sqrt{(-3)² + (-1)²}

                  = \sqrt{9 + 1} = \sqrt{10}

Distance PS = \sqrt{(1 - 4)² + (2 -5)²}

                  = \sqrt{(-3)² + (-3)²}

                  = \sqrt{9 + 9} =  3 \sqrt{2}

We see that the oposite sides are equal. Lets find the diagonal now,

Diagonal PR = \sqrt{(4 - 4)² + (3 -5)²}

                  = \sqrt{0 + 4} = 2

Diagonal QS = \sqrt{(1 - 7)² + (2 - 6)²}

                  = \sqrt{(-6)² + (-4)²}

                  = \sqrt{36 + 16} = \sqrt{52}

      We see that the diagonals are not equal.

So, clearly, the formed quadrilateral is a Parallelogram.

Class 10 Maths Ex 7.1 Question 7

Find the point on the x axis which is equidistant from (2, -5) and (-2, 9).

Solution: Let the point on the X axis be (x, 0)

Given, distance between (2, -5), (x, 0) = distance between (-2, 9), (x, 0)

⇒ \sqrt{(x - 2)²+(0 - (-5))²} = \sqrt{(x - (-2) )²+(0 - 9)²}

  ⇒ (x – 2)² + 5² = (x + 2)² + 9² [squaring both sides]

  ⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81

  ⇒ -4x -4x = 85 – 29

  ⇒ -8x = 56  ⇒ x = \cfrac {56} {-8}

  ⇒ x = -7

Therefore, The required point on the x axis is (-7, 0).

Class 10 Maths Ex 7.1 Question 8

Find the value of y for which the distance between the points P(2, -3) and Q(10, y) is 10 unites.

Solution: Given that,

Distance Between P(2, -3) and Q(10, y) = 10 unites

⇒ \sqrt{(10 - 2)² + (y - (-3) )²} = 10

⇒ (8)² + (y + 3)² = 100  [squaring both sides]

⇒ 64 + y² + 6y + 9 = 100

⇒ y² + 6y + 73 – 100 = 0

⇒ y² + 6y – 27 = 0

⇒ y² +9y – 3y – 27 = 0

⇒ y(y + 9) – 3(y + 9) = 0

⇒ (y + 9)(y – 3) = 0

So, y + 9 = 0 OR y – 3 = 0

⇒ y = – 9 OR y = 3

Therefore, Required y = -9 or 3

Class 10 Maths Ex 7.1 Question 9

If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), Find the value of x. Also find the distance QR and PR.

Solution: Given that,

Distance QR = Distance QP

⇒ \sqrt{(x - 0)² + (6 - 1 )²} = \sqrt{(5 - 0)² + (-3 - 1)²}

  ⇒ x² + 25 = 25 + 16

    ⇒ x² = 16

   So, x = ±√16 = ±4

   Now, For x = 4

QR = \sqrt{(4 - 0)² + (6 - 1 )²} = \sqrt{16 + 25} = \sqrt{41}

PR = \sqrt{(x - 5)² + (6 - (-3) )²}

     = \sqrt{(4 - 5)² + (6 +3 )²}

     = \sqrt{1 + 81} = \sqrt{82}

        For x = -4,

QR = \sqrt{(-4 - 0)² + (6 - 1 )²} = \sqrt{16 + 25}

     = \sqrt{41}

PR = \sqrt{(x - 5)² + (6 - (-3) )²}

     = \sqrt{(-4 - 5)² + (6 +3 )²}

     = \sqrt{81 + 81} = 9 \sqrt{2}

Therefore, Required value of x = ±4 and QR = √41 and PR = √82 or 9√2

Class 10 Maths Ex 7.1 Question 10

Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).

Solution: Given that,

Distance between (3, 6), (x, y) = Distance between (-3, 4), (x, y)

⇒ \sqrt{(x - 3)² + (y - 6)²} = \sqrt{(x - (-3))² + (y - 4)²}

⇒ (x – 3)² + (y – 6)² = (x +3)² + (y – 4)² [squaring both sides]

⇒ x² – 6x + 9 + y² -12y + 36 = x² + 6x + 9 + y² – 8y + 16

⇒ – 6x – 6x -12y + 8y + 45 – 25 = 0

⇒ – 12x – 4y + 20 = 0   [divide both side by -4]

⇒ 3x + y – 5 = 0

      — Which is the required condition.