Chapter 7: Coordinate Geometry

 Ans: Let P\[{\text{(x,y)}}\] be the required point.

Let A\[{\text{( - 1,7)}}\] and B\[{\text{(4, - 3)}}\] where \[{{\text{x}}_{\text{1}}}{\text{ =  - 1,}}{{\text{y}}_{\text{1}}}{\text{ = 7,}}{{\text{x}}_{\text{2}}}{\text{ = 4,}}{{\text{y}}_{\text{2}}}{\text{ =  - 3}}\] and \[{\text{m:n = 2:3}}\]

So, by section formula

\[{\text{P(x,y)}}\]= \[\left[ {\dfrac{{{\text{m}}{{\text{x}}_{\text{2}}}{\text{ +  n}}{{\text{x}}_{\text{1}}}}}{{{\text{m  +  n}}}}{\text{,}}\dfrac{{{\text{m}}{{\text{y}}_{\text{2}}}{\text{ +  n}}{{\text{y}}_{\text{1}}}}}{{{\text{m  +  n}}}}} \right]\]

Now, substituting the values we get,

\[{\text{x  =  }}\left[ {\dfrac{{{\text{2(4)  +  3( - 1)}}}}{{{\text{2  +  3}}}}} \right]{\text{ =  }}\dfrac{{{\text{8  -  3}}}}{{\text{5}}}{\text{  =  }}\dfrac{{\text{5}}}{{\text{5}}}{\text{  =  1}}\]

\[{\text{y  =  }}\left[ {\dfrac{{{\text{2( - 3) + 3(7)}}}}{{{\text{2 + 3}}}}} \right]{\text{  =  }}\dfrac{{{\text{ - 6  +  21}}}}{{\text{5}}}{\text{  = }}\dfrac{{{\text{15}}}}{{\text{5}}}{\text{  =  3}}\]

Therefore, the co-ordinates of point P are\[{\text{(1,3)}}\]

 Ans: 

Let line segment joining the points be A\[{\text{(4, - 1)}}\]  and B\[{\text{( - 2, - 3)}}\].

Let P\[{\text{(}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{)}}\] and Q\[{\text{(}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}\] are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB 

So, by section formula

\[{\text{P(x,y)}}\]= \[\left[ {\dfrac{{{\text{m}}{{\text{x}}_{\text{2}}}{\text{ +  n}}{{\text{x}}_{\text{1}}}}}{{{\text{m  +  n}}}}{\text{,}}\dfrac{{{\text{m}}{{\text{y}}_{\text{2}}}{\text{ +  n}}{{\text{y}}_{\text{1}}}}}{{{\text{m  +  n}}}}} \right]\]

Therefore, by observation point P divides AB internally in the ratio \[{\text{1:2}}\]

Now, substituting the values we get,

\[{{\text{x}}_1}{\text{ =  }}\left[ {\dfrac{{{\text{1( - 2) + 2(4)}}}}{{1 + 2}}} \right]{\text{  =  }}\dfrac{{{\text{ -  2  +  8}}}}{3}{\text{  =  }}\dfrac{6}{3}{\text{  =  2}}\]

\[{{\text{y}}_1}{\text{ =  }}\left[ {\dfrac{{{\text{1( - 3) + 2( - 1)}}}}{{{\text{1 + 2}}}}} \right]{\text{  =  }}\dfrac{{{\text{ - 3 - 2}}}}{3}{\text{  =  }}\dfrac{{{\text{ - 5}}}}{3}\]

Therefore, P\[{\text{(}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{)}}\]=\[{\text{(2, - }}\dfrac{{\text{5}}}{{\text{3}}}{\text{)}}\]

Therefore, by observation point Q divides AB internally in the ratio \[{\text{2:1}}\]

Now, substituting the values we get,

\[{{\text{x}}_1}{\text{ = }}\left[ {\dfrac{{{\text{2( - 2) + 1(4)}}}}{{2 + 1}}} \right]{\text{  =  }}\dfrac{{{\text{ - 4 + 4}}}}{3}{\text{  =  }}\dfrac{0}{3}{\text{  =  0}}\]

\[{{\text{y}}_1}{\text{ =  }}\left[ {\dfrac{{{\text{2( - 3) + 1( - 1)}}}}{{2 + 1}}} \right]{\text{ = }}\dfrac{{{\text{ - 6 - 1}}}}{3}{\text{ = }}\dfrac{{{\text{ - 7}}}}{3}\]

Therefore, Q\[{\text{(}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}\]= \[{\text{(0, - }}\dfrac{7}{{\text{3}}}{\text{)}}\]

Ans: By observation, that Niharika posted the green flag at of the distance P i.e., \[{\text{(}}\dfrac{{\text{1}}}{{\text{4}}}{ \times\text{ 100)}}\;{\text{m = 25}}\;{\text{m}}\]  from the starting point of 2nd line. Therefore, the coordinates of this point P is\[{\text{(2,25)}}\]

Similarly, Preet posted red flag at of the distance Q i.e., \[{\text{(}}\dfrac{{\text{1}}}{5}{ \times \text{100)}}\;{\text{m = 20}}\;{\text{m}}\]     from the starting point of 8th line. Therefore, the coordinates of this point Q are\[{\text{(8,20)}}\]

We know that the distance between the two points is given by the distance formula

i.e. \[\sqrt {{{{\text{(}}{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}{\text{)}}}^{\text{2}}}{\text{  +  (}}{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}} \]

To find the distance between these flags PQ by substituting the values we get,

PQ =\[\sqrt {{{{\text{(8 - 2)}}}^{\text{2}}}{\text{ +  (25 - 20}}{{\text{)}}^{\text{2}}}} {\text{ =  }}\sqrt {{\text{36  +  25}}} {\text{   =  }}\sqrt {{\text{61}}} \;{\text{m}}\]

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be M\[{\text{(x,y)}}\]

So, by section formula

M\[{\text{(x,y)}}\]= \[\left[ {\dfrac{{{\text{m}}{{\text{x}}_{\text{2}}}{\text{ +  n}}{{\text{x}}_{\text{1}}}}}{{{\text{m  +  n}}}}{\text{,}}\dfrac{{{\text{m}}{{\text{y}}_{\text{2}}}{\text{ +  n}}{{\text{y}}_{\text{1}}}}}{{{\text{m  +  n}}}}} \right]\]

Now, substituting the values we get,

\[{\text{x = }}\dfrac{{{\text{2 + 8}}}}{2}{\text{ = }}\dfrac{{10}}{2}{\text{ = 5}}\]

\[{\text{y = }}\dfrac{{25 + 20}}{2}{\text{ = }}\dfrac{{{\text{45}}}}{2}{\text{ =  22}}{\text{.5}}\]

Therefore, Rashmi should post her blue flag at \[{\text{22}}{\text{.5 m}}\] on fifth line.

 Ans:

Let the ratio in which the line segment joining A\[{\text{( - 3,10)}}\] and B\[{\text{(6, - 8)}}\] is divided by point P\[{\text{( - 1,6)}}\] be \[{\text{k:1}}\].

So, by section formula

M\[{\text{(x,y)}}\]= \[\left[ {\dfrac{{{\text{m}}{{\text{x}}_{\text{2}}}{\text{ +  n}}{{\text{x}}_{\text{1}}}}}{{{\text{m  +  n}}}}{\text{,}}\dfrac{{{\text{m}}{{\text{y}}_{\text{2}}}{\text{ +  n}}{{\text{y}}_{\text{1}}}}}{{{\text{m  +  n}}}}} \right]\]

\[\therefore \;{\text{ - 1 = }}\dfrac{{{\text{6k - 3}}}}{{{\text{k + 1}}}}\]

\[{\text{ - k - 1 = 6k - 3}}\]

\[{\text{7k = 2}}\]                             (by cross multiplying and transposing)

\[{\text{k = }}\dfrac{2}{7}\]

Hence the point P divides AB in the ratio\[2:7\]

 Ans:

 Let the ratio be \[{\text{k:1}}\]and let the line segment joining A\[(1, - 5)\] and B\[( - 4,5)\]

So, by section formula

M\[{\text{(x,y)}}\]= \[\left[ {\dfrac{{{\text{m}}{{\text{x}}_{\text{2}}}{\text{ +  n}}{{\text{x}}_{\text{1}}}}}{{{\text{m  +  n}}}}{\text{,}}\dfrac{{{\text{m}}{{\text{y}}_{\text{2}}}{\text{ +  n}}{{\text{y}}_{\text{1}}}}}{{{\text{m  +  n}}}}} \right]\]

Now, by substituting the values we get the coordinate of the point of division \[{\text{[}}\dfrac{{{\text{ - 4k + 1}}}}{{{\text{k + 1}}}}{\text{,}}\dfrac{{{\text{5k - 5}}}}{{{\text{k + 1}}}}{\text{]}}\]

We know that y-coordinate of any point on x-axis is \[{\text{0}}\] \[\therefore {\text{ }}\dfrac{{{\text{5k - 5}}}}{{{\text{k + 1}}}} = 0 \Rightarrow {\text{5k}} - 5 = 0 \Rightarrow {\text{5k}} = 5 \Rightarrow {\text{k}} = 1\]

Therefore, x-axis divides it in the ratio \[1:1\]

Division point = \[{\text{(}}\dfrac{{{\text{ - 4(1) + 1}}}}{{{\text{1 + 1}}}}{\text{,}}\dfrac{{{\text{5(1) + 5}}}}{{{\text{1 + 1}}}}{\text{) = (}}\dfrac{{{\text{ - 4 + 1}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + 5}}}}{{\text{2}}}{\text{) = (}}\dfrac{{{\text{ - 3}}}}{{\text{2}}}{\text{,0)}}\]

 Ans: 

Let A\[{\text{(1,2)}}\], B\[{\text{(4, y)}}\],C\[{\text{(x,6)}}\] and D\[{\text{(3,5)}}\] are the vertices of a parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, Intersection point O of diagonal AC and BD also divides these diagonals Therefore, O is the mid-point of AC and BD. 

If O is the mid-point of AC, then the coordinates of O are

\[{\text{(}}\dfrac{{{\text{1 + x}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{2 + 6}}}}{{\text{2}}}{\text{)}} \Rightarrow {\text{(}}\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{,4)}}\]

If O is the mid-point of BD, then the coordinates of O are\[{\text{(}}\dfrac{{{\text{4 + 3}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{)}} \Rightarrow {\text{(}}\dfrac{{\text{7}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{)}}\]

Since both the coordinates are of the same point O, \[\dfrac{{{\text{1 + x}}}}{{\text{2}}}{\text{ = }}\dfrac{{\text{7}}}{{\text{2}}}\;{\text{and}}\;\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{ = 4}}\]

\[{\text{x  + 1  =  7}}\;{\text{and}}\;{\text{5  +  y  =  8}}\]

\[{\text{x  =  6}}\;{\text{and}}\;{\text{y  =  3}}\]

 Ans: 

Let the coordinates of point A be\[{\text{(x,y)}}\] 

Mid-point of AB is C\[{\text{(2, - 3)}}\], which is the center of the circle.

\[{\text{(2, - 3) = (}}\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{y + 4}}}}{{\text{2}}}{\text{)}}\]

\[\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{ = 2 and}}\;\dfrac{{{\text{y + 4}}}}{{\text{2}}} =  - 3\]

\[{\text{x  +  1 =  4  and}}\;{\text{y  +  4  =   - 6  }}\]

\[{\text{x = 3 and}}\;{\text{y =  - 10}}\]. 

So, coordinates of A are \[{\text{(3, - 10)}}\]

 Ans: 

The coordinates of point A and B are\[{\text{( - 2, - 2)}}\] and\[{\text{(2, - 4)}}\] respectively and \[{\text{AP = }}\dfrac{{\text{3}}}{{\text{7}}}{\text{AB}}\] so, \[\dfrac{{{\text{AP}}}}{{{\text{AB}}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{7}}}\]

We know that AB = AP + PB from figure, 

\[\dfrac{{{\text{AP + PB}}}}{{{\text{AP}}}}{\text{ = }}\dfrac{{{\text{3 + 4}}}}{3}\]

\[1 + \dfrac{{{\text{PB}}}}{{{\text{AP}}}}{\text{ = 1 + }}\dfrac{{\text{4}}}{3}\]

\[\dfrac{{{\text{PB}}}}{{{\text{AP}}}}{\text{ = }}\dfrac{{\text{4}}}{3}\]

Therefore, \[{\text{PB:AP = 4:3}}\]

Point P\[{\text{(x,y)}}\] divides the line segment AB in the ratio\[{\text{3:4}}\]

Coordinates of P\[{\text{(x,y)}}\] = \[{\text{(}}\dfrac{{{\text{3 * 2 + 4 * ( - 2)}}}}{{{\text{3 + 4}}}}{\text{,}}\dfrac{{{\text{3 * ( - 4) + 4 * ( - 2)}}}}{{{\text{3 + 4}}}}{\text{)}}\]

                                           = \[{\text{(}}\dfrac{{{\text{6 - 8}}}}{{\text{7}}}{\text{,}}\dfrac{{{\text{ - 12 - 8}}}}{{\text{7}}}{\text{)}}\;{\text{ = }}\;{\text{(}}\dfrac{{{\text{ - 2}}}}{{\text{7}}}{\text{,}}\dfrac{{{\text{ - 20}}}}{{\text{7}}}{\text{)}}\]

 Ans:

By observation, \[{{\text{P}}_1}\],\[{{\text{P}}_2}\],\[{{\text{P}}_3}\] that points divides the line segment A\[{\text{( - 2,2)}}\]and B\[{\text{(2,8)}}\]into four equal parts. Point\[{{\text{P}}_1}\] divides the line segment A\[{{\text{P}}_2}\] into two equal parts 

Hence, Coordinates of \[{{\text{P}}_1}\]= \[(\dfrac{{1 \times 2 + 3 \times ( - 2)}}{{1 + 3}},\dfrac{{1 \times 8 + 3 \times 2}}{{1 + 3}})\; = \;( - 1,\dfrac{7}{2})\]

Point\[{{\text{P}}_2}\] divides the line segment AB into two equal parts

Coordinates of \[{{\text{P}}_2}\]= \[(\dfrac{{2 + ( - 2)}}{2},\dfrac{{2 + 8}}{2})\; = \;(0,5)\]

Point \[{{\text{P}}_{\text{3}}}\] divides the line segment B \[{{\text{P}}_2}\] into two equal parts

Coordinates of \[{{\text{P}}_3}\]= \[(\dfrac{{3 \times 2 + 1 \times ( - 2)}}{{1 + 3}},\dfrac{{3 \times 8 + 1 \times 2}}{{1 + 3}})\; = \;(1,\dfrac{{13}}{2})\]

 Ans: 

Let A\[{\text{(3,0)}}\],B\[{\text{(4,5)}}\],C\[{\text{( - 1,4)}}\] and D\[{\text{( - 2, - 1)}}\] are the vertices of a rhombus ABCD. 

We know that the distance between the two points is given by the distance formula

i.e. \[\sqrt {{{{\text{(}}{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}{\text{)}}}^{\text{2}}}{\text{ + (}}{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}} \]

Therefore, distance between A\[{\text{(3,0)}}\]and C\[{\text{( - 1,4)}}\] is given by

Length of diagonal AC = \[\sqrt {{{[3 - ( - 1)]}^2} + {{(0 - 4)}^2}}  = \sqrt {16 + 16}  = 4\sqrt 2 \]

Therefore, distance between B\[{\text{(4,5)}}\]and D \[{\text{( - 2, - 1)}}\] is given by

Length of diagonal BD = \[\sqrt {{{[4 - ( - 2)]}^2} + {{(5 - ( - 1))}^2}}  = \sqrt {36 + 36}  = 6\sqrt 2 \]

Area of rhombus ABCD = \[\dfrac{{\text{1}}}{{\text{2}}} \times{\text{ (product}}\;{\text{of}}\;{\text{lengths}}\;{\text{of}}\;{\text{diagonals)}}\]

                                      = \[\dfrac{{\text{1}}}{{\text{2}}}{\text{ AC}} \times {\text{BD}}\]

                                      = \[\dfrac{{\text{1}}}{{\text{2}}} \times{\text{ 4}}\sqrt {\text{2}}  \times{\text{6}}\sqrt {\text{2}} {\text{ = 24}}\;{\text{square}}\;{\text{units}}\]