NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry - Exercise 7.1

1. Find the distance between the following pairs of points:

(i) \[\left( \text{2,3} \right)\text{,}\left( \text{4,1} \right)\] 

Ans: The given points are \[\text{A}\left( \text{2, 3} \right)\] and \[\text{B}\left( \text{4, 1} \right)\]

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]  

By comparing the points with the formula we get, \[{{\text{x}}_{\text{1}}}\text{=2}\], \[{{\text{y}}_{\text{1}}}\text{=3}\], \[{{\text{x}}_{\text{2}}}\text{=4}\], \[{{\text{y}}_{\text{2}}}\text{=1}\]

Distance between \[\text{A}\left( \text{2,3} \right)\] and \[\text{B}\left( \text{4,1} \right)\] is given by,

AB = \[\sqrt{{{\text{(2-4)}}^{\text{2}}}\text{+(3-1}{{\text{)}}^{\text{2}}}}\]

      = \[\sqrt{{{\text{(-2)}}^{\text{2}}}\text{+(2}{{\text{)}}^{\text{2}}}}\]

      = \[\sqrt{\text{4+4}}\]

      = \[\sqrt{\text{8}}\]

      = \[\text{2}\sqrt{\text{2}}\]

(ii) \[\left( -\mathbf{5},\mathbf{7} \right),\text{ }\left( -\mathbf{1},\mathbf{3} \right)\]

Ans: The given points are \[\text{C}\left( \text{-5, 7} \right)\] and \[\text{D}\left( \text{-1, 3} \right)\]

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]  

By comparing the points with the formula we get, \[{{\text{x}}_{\text{1}}}\text{= -5}\], \[{{\text{y}}_{\text{1}}}\text{= 7}\], \[{{\text{x}}_{\text{2}}}\text{= -1}\], \[{{\text{y}}_{\text{2}}}\text{= 3}\]

Distance between \[C\left( \text{-5,7} \right)\]and\[D\left( \text{-1,3} \right)\] is given by,

CD  = \[\sqrt{{{\text{(-5-(-1))}}^{\text{2}}}\text{+(7-3}{{\text{)}}^{\text{2}}}}\]

       = \[\sqrt{{{\text{(-4)}}^{\text{2}}}\text{+(4}{{\text{)}}^{\text{2}}}}\]

       = \[\sqrt{\text{16+16}}\]

       = \[\sqrt{\text{32}}\]

       = \[\text{4}\sqrt{\text{2}}\]

(iii) \[\left( \mathbf{a},\mathbf{b} \right),\left( -\mathbf{a},-\mathbf{b} \right)\]

Ans: The given points are \[\text{X}\left( \text{a, b} \right)\] and \[\text{Y}\left( \text{-a, -b} \right)\]

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]  

By comparing the points with the formula we get, \[{{\text{x}}_{\text{1}}}\text{= a}\], \[{{\text{y}}_{\text{1}}}\text{= b}\], \[{{\text{x}}_{\text{2}}}\text{= -a}\],\[{{\text{y}}_{\text{2}}}\text{= -b}\] 

Distance between \[\text{X}\left( \text{a, b} \right)\] and \[\text{Y}\left( \text{-a, -b} \right)\] is given by, 

XY = \[\sqrt{{{\text{(a-(-a))}}^{\text{2}}}\text{+(b-(-b)}{{\text{)}}^{\text{2}}}}\]

            = \[\sqrt{{{\text{(2a)}}^{\text{2}}}\text{+(2b}{{\text{)}}^{\text{2}}}}\]

           = \[\sqrt{\text{4}{{\text{a}}^{\text{2}}}\text{+4}{{\text{b}}^{\text{2}}}}\]

           = \[\text{2}\sqrt{{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}}\]

2. Find the distance between the points \[\left( \text{0,0} \right)\] and \[\left( \text{36,15} \right)\]. Can you now find the distance between the two towns A and B discussed in Section 7.2? 

Ans: According to the , the given points are \[\text{P}\left( \text{0,0} \right)\] and \[\text{Q}\left( \text{36,15} \right)\].

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]  

By comparing the points with the formula we get, \[{{\text{x}}_{\text{1}}}\text{=0}\], \[{{\text{y}}_{\text{1}}}\text{=0}\], \[{{\text{x}}_{\text{2}}}\text{=36}\], \[{{\text{y}}_{\text{2}}}\text{=15}\].

The distance between the points \[\text{P}\left( \text{0,0} \right)\] and \[\text{Q}\left( \text{36,15} \right)\] is given by,

PQ  = \[\sqrt{{{\text{(36-0)}}^{\text{2}}}\text{+(15-0}{{\text{)}}^{\text{2}}}}\]

       = \[\sqrt{\text{3}{{\text{6}}^{\text{2}}}\text{+1}{{\text{5}}^{\text{2}}}}\]

       = \[\sqrt{\text{1296+225}}\]

       = \[\sqrt{\text{1521}}\]

       = \[\text{39}\]

Yes, we can find the distance between two towns A and B using the distance formula only. Two town are situated at point  (0,0) and (36,15) respectively. So, the distance between both town will be 39 km.

3. Determine if the points \[\left( \text{1,5} \right)\text{, }\left( \text{2,3} \right)\] and \[\left( \text{-2,-11} \right)\] are collinear.

Ans: Three points are collinear if and only if they are on the same line. 

Let the given points are \[A\left( \text{1,5} \right)\text{, B}\left( \text{2,3} \right)\] and \[C\left( \text{-2,-11} \right)\] 

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\],

\[AB=\sqrt{{{(1-2)}^{2}}+{{(5-3)}^{2}}}\]

          = \[\sqrt{{{\text{1}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}}\]

          = \[\sqrt{\text{5}}\]

\[B\text{C}=\sqrt{{{(2-(-2))}^{2}}+{{(3-(-11))}^{2}}}\]

          = \[\sqrt{\text{16+196}}\]

         = \[\sqrt{\text{212}}\]

\[\text{AC}=\sqrt{{{(1-(-2))}^{2}}+{{(5-(-11))}^{2}}}\]

           = \[\sqrt{\text{9+256}}\]

           = \[\sqrt{265}\]

Here,\[\text{AB+AC}\ne \text{BC}\] and \[\text{BC+AC}\ne AB\] and \[\text{AB+BC}\ne A\text{C}\]. Since, the sum of distance of any two point is not equal to third point.

Therefore, the points \[\left( \text{1,5} \right)\text{, }\left( \text{2,3} \right)\] and \[\left( \text{-2,-11} \right)\] are not collinear.

4. Check whether \[\left( \text{5,-2} \right)\text{, }\left( \text{6,4} \right)\] and \[\left( \text{7,-2} \right)\] are the vertices of an isosceles triangle.

Ans: To prove the points to be an isosceles triangle the points must be collinear and any two sides must be of equal length.  

Let the points \[\left( \text{5,-2} \right)\text{, }\left( \text{6,4} \right)\] and \[\left( \text{7,-2} \right)\] are representing the vertices

A, B, and C of the given triangle respectively.

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\],

\[AB=\sqrt{{{(5-6)}^{2}}+{{(-2-4)}^{2}}}\]

           = \[\sqrt{{{\text{(-1)}}^{2}}\text{+(-6}{{\text{)}}^{2}}}\]

           = \[\sqrt{\text{1+36}}\]

           = \[\sqrt{37}\]

\[BC=\sqrt{{{(6-7)}^{2}}+{{(4-(-2))}^{2}}}\]

           = \[\sqrt{{{\text{(-1)}}^{2}}\text{+(6}{{\text{)}}^{2}}}\]

           = \[\sqrt{\text{1+36}}\]

           = \[\sqrt{37}\]

\[AC=\sqrt{{{(5-7)}^{2}}+{{(-2-(-2))}^{2}}}\]

           = \[\sqrt{{{\text{(-2)}}^{2}}\text{+(0}{{\text{)}}^{2}}}\]

           = \[\sqrt{4}\]

           = 2

Here, AB = BC. As two sides are equal in length and the sum of distance of any two point is not equal to third point. Therefore, ABC is an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Ans: From the above figure, the coordinates of A, B, C, D are\[\left( \text{3,4} \right)\text{, }\left( \text{6,7} \right)\text{, }\left( \text{9,4} \right)\] and \[\left( \text{6,1} \right)\] respectively.

Using the Distance formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]

\[AB=\sqrt{{{(3-6)}^{2}}+{{(4-7)}^{2}}}\]

           = \[\sqrt{{{\text{(-3)}}^{2}}\text{+(-3}{{\text{)}}^{2}}}\]

           = \[\sqrt{\text{9+9}}\]

           = \[\sqrt{18}\]

           = \[3\sqrt{2}\]

\[BC=\sqrt{{{(6-9)}^{2}}+{{(7-4)}^{2}}}\]

          = \[\sqrt{{{\text{(-3)}}^{2}}\text{+}{{\text{3}}^{2}}}\]

          = \[\sqrt{9+9}\]

          = \[\sqrt{18}\]

          =\[3\sqrt{2}\]

\[CD=\sqrt{{{(9-6)}^{2}}+{{(4-1)}^{2}}}\]

          = \[\sqrt{{{\text{3}}^{2}}\text{+}{{\text{3}}^{2}}}\]

          = \[\sqrt{9+9}\]

          = \[\sqrt{18}\]

        = \[3\sqrt{2}\]

\[AD=\sqrt{{{(3-6)}^{2}}+{{(4-1)}^{2}}}\]

        = \[\sqrt{{{\text{(-3)}}^{2}}\text{+}{{\text{3}}^{2}}}\]

       = \[\sqrt{9+9}\]

       = \[\sqrt{18}\]

       = \[3\sqrt{2}\]

Diagonal \[AC=\sqrt{{{(3-9)}^{2}}+{{(4-4)}^{2}}}\]

                         = \[\sqrt{{{\text{(-6)}}^{2}}\text{+}{{\text{0}}^{2}}}\]

                         = \[\text{6}\]

Diagonal \[BD=\sqrt{{{(6-6)}^{2}}+{{(7-1)}^{2}}}\]

                        = \[\sqrt{{{0}^{2}}\text{+}{{\text{6}}^{2}}}\]

                        = \[\text{6}\]

Here, all four sides AB, BC, CD, and AD are of same length and diagonals AC and BD are also of equal length . Therefore, we can conclude that ABCD is a square and hence, Champa was correct.

6. Name the type of quadrilateral formed, if any, by the following points, and 

give reasons for your Ans:

i. \[\text{(-1,-2), (1,0), (-1,2), (-3,0)}\]

Ans: Let the points \[\text{(-1,-2), (1,0), (-1,2), (-3,0)}\]be representing the vertices A, B, C, and D of the given quadrilateral respectively.      

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\],

\[AB=\sqrt{{{(-1-1)}^{2}}+{{(-2-0)}^{2}}}\]

   \[=\sqrt{{{(-2)}^{2}}\text{+(-2}{{\text{)}}^{2}}}\]

   \[=\sqrt{\text{4+4}}\]

   \[=\sqrt{8}\text{ }\]

         = \[\text{2}\sqrt{2}\]

\[BC=\sqrt{{{(1-(-1))}^{2}}+{{(0-2)}^{2}}}\]

   \[=\sqrt{{{(2)}^{2}}\text{+(-2}{{\text{)}}^{2}}}\]

       \[=\sqrt{\text{4+4}}\]

       \[=\sqrt{8}\]

       = \[\text{2}\sqrt{2}\]

\[CD=\sqrt{{{(-1-(-3))}^{2}}+{{(2-0)}^{2}}}\]

   \[=\sqrt{{{2}^{2}}\text{+}{{\text{2}}^{2}}}\]

      \[=\sqrt{\text{4+4}}\]

      \[=\sqrt{8}\]

      = \[\text{2}\sqrt{2}\]

\[AD=\sqrt{{{(-1-(-3))}^{2}}+{{(-2-0)}^{2}}}\]

   \[=\sqrt{{{2}^{2}}\text{+}{{\text{2}}^{2}}}\]

       \[=\sqrt{\text{4+4}}\]

       \[=\sqrt{8}\]

       = \[\text{2}\sqrt{2}\]

Diagonal \[AC=\sqrt{{{(-1-(-1))}^{2}}+{{((-2)-2)}^{2}}}\]

            \[=\sqrt{{{0}^{{}}}\text{+}{{\text{4}}^{2}}}\]

            \[=\sqrt{16}\]

            \[\text{= 4}\]

Diagonal \[BD=\sqrt{{{(1-(-3))}^{2}}+{{(0-0)}^{2}}}\]

     \[=\sqrt{{{\text{4}}^{2}}+0}\]

                 \[=\sqrt{16}\]

                     \[\text{= 4}\]

Therefore, AB = BC = CD = AD

Here, all the sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

ii. \[\text{(-3,5), (3,1), (0,3), (-1,-4)}\]

Ans: Let the points \[\text{(-3,5), (3,1), (0,3), (-1,-4)}\] be representing the vertices A, B, C, and D of the given quadrilateral respectively.      

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\],

\[AB=\sqrt{{{(-3-3)}^{2}}+{{(5-1)}^{2}}}\]

\[=\sqrt{{{(-6)}^{2}}\text{+(4}{{\text{)}}^{2}}}\]

\[=\sqrt{36+16}\]

\[=\sqrt{52}\]

       \[\text{=2}\sqrt{13}\]

\[BC=\sqrt{{{(3-0)}^{2}}+{{(1-3)}^{2}}}\]

  \[=\sqrt{{{3}^{2}}\text{+(-2}{{\text{)}}^{2}}}\]

\[=\sqrt{\text{9+4}}\]

\[=\sqrt{13}\]

\[CD=\sqrt{{{(0-(-1))}^{2}}+(3-{{(-4)}^{2}}}\]

  \[=\sqrt{{{1}^{2}}\text{+}{{\text{7}}^{2}}}\]

\[=\sqrt{\text{1+49}}\]

\[=\sqrt{50}\]

      \[\text{=5}\sqrt{2}\]

\[AD=\sqrt{{{(-3-(-1))}^{2}}+(5-{{(-4)}^{2}}}\]

  \[=\sqrt{{{(-2)}^{2}}\text{+}{{\text{9}}^{2}}}\]

\[=\sqrt{\text{4+81}}\]

\[=\sqrt{85}\]

Diagonal \[AC=\sqrt{{{(0-(-3))}^{2}}+{{(3-5)}^{2}}}\]

                       \[=\sqrt{{{3}^{2}}\text{+(-2}{{\text{)}}^{2}}}\]

            \[=\sqrt{\text{9+4}}\]

            \[=\sqrt{13}\]

Diagonal \[BD=\sqrt{{{(-1-3)}^{2}}+{{(-4-1)}^{2}}}\]

                       \[=\sqrt{{{(-4)}^{2}}+{{(-5)}^{2}}}\]

               \[=\sqrt{16+25}\]

                   \[\text{= }\sqrt{41}\]

Here, all the sides of this quadrilateral are of different length i.e.,\[\text{AB}\ne \text{BC}\ne \text{AC}\ne \text{AD}\] and also AC + BC = AB which means that A, B, C are collinear. Therefore, no quadrilateral can be formed from these points.

iii. \[\text{(4,5), (7,6), (4,3), (1,2)}\]

Ans: Let the points \[\text{(4,5), (7,6), (4,3), (1,2)}\] be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\],

\[AB=\sqrt{{{(4-7)}^{2}}+{{(5-6)}^{2}}}\]

  \[=\sqrt{{{(-3)}^{2}}\text{+(-1}{{\text{)}}^{2}}}\]

  \[=\sqrt{9+1}\]

  \[=\sqrt{10}\]

\[BC=\sqrt{{{(7-4)}^{2}}+{{(6-3)}^{2}}}\]

  \[=\sqrt{{{3}^{2}}\text{+}{{\text{3}}^{2}}}\]

  \[=\sqrt{\text{9+9}}\]

  \[=\sqrt{18}\]

\[CD=\sqrt{{{(4-1)}^{2}}+{{(3-2)}^{2}}}\]

     \[=\sqrt{{{3}^{2}}\text{+}{{\text{1}}^{2}}}\]

   \[=\sqrt{\text{9+1}}\]

   \[=\sqrt{10}\]

\[AD=\sqrt{{{(4-1)}^{2}}+{{(5-2)}^{2}}}\]

\[=\sqrt{{{3}^{2}}\text{+}{{\text{3}}^{2}}}\]

\[=\sqrt{\text{9+9}}\]

\[=\sqrt{18}\]

Diagonal \[AC=\sqrt{{{(4-4)}^{2}}+{{(5-3)}^{2}}}\]

\[=\sqrt{0\text{+}{{\text{2}}^{2}}}\]

\[=\sqrt{4}\text{  = 2}\]

Diagonal \[BD=\sqrt{{{(7-1)}^{2}}+{{(6-2)}^{2}}}\]

   \[=\sqrt{{{6}^{2}}+{{4}^{2}}}\]

   \[=\sqrt{36+16}\text{  = }\sqrt{52}\]

   \[=2\sqrt{13}\]

Here, the opposite sides of this quadrilateral are of the same length i.e., AB = CD and BC = AD. But, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

7. Find the point on the x-axis which is equidistant from \[\text{(2,-5)}\] and \[\text{(-2,9)}\]. 

Ans: If the point is on x-axis the co-ordinates will be \[P\left( \text{x, 0} \right)\]. We have to find a point on x-axis which is equidistant from \[A\left( \text{2,-5} \right)\] and \[B\left( \text{-2,9} \right)\].

Using the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]

\[PA=\sqrt{{{(x-2)}^{2}}+{{(0-(-5))}^{2}}}\]

       \[=\sqrt{{{(x-2)}^{2}}+{{5}^{2}}}\]

\[PB=\sqrt{{{(x-(-2))}^{2}}+{{(0-(-9))}^{2}}}\]

   \[=\sqrt{{{(x+2)}^{2}}+{{9}^{2}}}\]

By the given condition, these distances are equal in measure.

Hence PA = PB

\[\sqrt{{{(x-2)}^{2}}+{{5}^{2}}}=\sqrt{{{(x+2)}^{2}}+{{9}^{2}}}\]

\[\Rightarrow {{(x-2)}^{2}}+25={{(x+2)}^{2}}+81\] 

\[\Rightarrow {{\text{x}}^{2}}+4-4x+25={{x}^{2}}+4+4x+81\] 

\[\Rightarrow \text{8x= 25-81}\]

\[\Rightarrow \text{8x= -56}\]

\[\Rightarrow \text{x= -7}\]

Hence, \[P\left( \text{-7, 0} \right)\] is the point on the x-axis which is equidistance from the given points.

8. Find the values of y for which the distance between the points \[\text{P(2,-3)}\] and \[\text{Q(10,y)}\] is \[\text{10}\] units.

Ans: We know that the distance between the two points is given by the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]

Distance between points \[A\left( \text{2,-3} \right)\] and \[B\left( \text{10,y} \right)\] is \[\text{10}\] units. Therefore, by substituting the values of points A (2, −3) and B (10, y) in Distance Equation

\[\sqrt{{{\text{(2-10)}}^{\text{2}}}\text{+(-3-y}{{\text{)}}^{\text{2}}}}=10\] 

\[\Rightarrow \sqrt{{{\text{(-8)}}^{\text{2}}}\text{+(3+y}{{\text{)}}^{\text{2}}}}=10\]

Squaring both sides

\[\Rightarrow \text{64+(3+y}{{\text{)}}^{\text{2}}}=100\]

\[\Rightarrow {{\text{(y+3)}}^{\text{2}}}=36\]

\[\Rightarrow \text{(y+3)}=\pm 6\]

\[\Rightarrow \text{(y+3)}=6\] or \[\text{(y+3)}=-6\]

Therefore, \[\text{y = -3 or 9}\] are the possible values for y.

9. If \[\text{Q(0,1)}\] is equidistant from \[\text{P(5,-3)}\] and \[\text{R(x,6)}\], find the values of x. Also find the distance QR and PR.

Ans: We know that the distance between the two points is given by the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]

As \[\text{Q(0,1)}\] is equidistant from \[\text{P(5,-3)}\] and \[\text{R(x,6)}\]. 

So, PQ = QR

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

\[\Rightarrow \sqrt{{{5}^{2}}+{{(-4)}^{2}}}=\sqrt{{{(-x)}^{2}}+{{(-5)}^{2}}}\]

\[\Rightarrow 25+16={{x}^{2}}+25\]      (By squaring both sides)

\[\Rightarrow 16={{x}^{2}}\]

\[\Rightarrow {{x}^{2}}=\pm 4\]

Therefore, point R is \[\text{(4,6)}\] or \[\text{(-4,6)}\]

Case (1),

When point R is \[\text{(4,6)}\], 

Distance between \[\text{P(5,-3)}\] and \[\text{R(4,6)}\]

PR=\[\sqrt{{{(5-4)}^{2}}+{{(-3-6)}^{2}}}\]

     \[=\sqrt{{{(1)}^{2}}+{{(-9)}^{2}}}\]

     \[=\sqrt{1+81}\]

      \[=\sqrt{82}\]

Distance between \[\text{Q(0,1)}\] and \[\text{R(4,6)}\]

QR = \[\sqrt{{{(0-4)}^{2}}+{{(1-6)}^{2}}}\]

      \[=\sqrt{{{(-4)}^{2}}+{{(-5)}^{2}}}\]

      \[=\sqrt{16+25}\] 

      \[=\sqrt{41}\]

Case (2), 

When point R is \[\text{(-4,6)}\], 

Distance between \[\text{P(5,-3)}\] and \[\text{R(-4,6)}\]

PR = \[\sqrt{{{(5-(-4))}^{2}}+{{(-3-6)}^{2}}}\]

      \[=\sqrt{{{(9)}^{2}}+{{(-9)}^{2}}}\]

      \[=\sqrt{81+81}\]

       \[=9\sqrt{2}\]

Distance between \[\text{Q(0,1)}\] and \[\text{R(-4,6)}\]

QR =\[\sqrt{{{(0-(-4))}^{2}}+{{(1-6)}^{2}}}\]

       \[=\sqrt{{{(4)}^{2}}+{{(-5)}^{2}}}\]

       \[=\sqrt{16+25}\]

        \[=\sqrt{41}\]

10. Find a relation between x and y such that the point (x, y) is equidistant from the point \[\text{(3,6)}\] and \[\text{(-3,4)}\].

Ans: We know that the distance between the two points is given by the Distance Formula = \[\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}\]

Point P(x, y) be equidistant from points \[\text{A(3,6)}\] and \[\text{B(-3,4)}\].

Therefore, PA = PB

\[\sqrt{{{(x-3)}^{2}}+{{(y-6)}^{2}}}=\sqrt{{{(x-(-3))}^{2}}+{{(y-4)}^{2}}}\]

\[\Rightarrow \sqrt{{{(x-3)}^{2}}+{{(y-6)}^{2}}}=\sqrt{{{(x+3)}^{2}}+{{(y-4)}^{2}}}\]

Squaring both sides

\[\Rightarrow {{(x-3)}^{2}}+{{(y-6)}^{2}}={{(x+3)}^{2}}+{{(y-4)}^{2}}\]

\[\Rightarrow {{x}^{2}}+9-6x+{{y}^{2}}+36-12y={{x}^{2}}+9+6x+{{y}^{2}}+16-8y\]

\[\Rightarrow 36-16=6x+6x+12y-8y\]

\[\Rightarrow 20=12x+4y\]

\[\Rightarrow 3x+y=5\]

\[\Rightarrow 3x+y-5=0\]

Thus, the relation between x and y is given by \[3x+y-5=0\]

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

Opting for the NCERT solutions for Ex 7.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Coordinate Geometry textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 7 Exercise 7.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

NCERT Solutions for Class 10 Maths Chapter 7 All Other Exercises