Chapter 7: Coordinate Geometry

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

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

      = \[\sqrt{\text{4+4}}\]

      = \[\sqrt{\text{8}}\]

      = \[\text{2}\sqrt{\text{2}}\]

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

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

       = \[\sqrt{\text{16+16}}\]

       = \[\sqrt{\text{32}}\]

       = \[\text{4}\sqrt{\text{2}}\]

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

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

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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

       \[=\sqrt{16+25}\]

        \[=\sqrt{41}\]

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