NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Ans : (i) distance between the two points is given by
\( \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
Therefore. distance between (2, 3) and (4, 1) is given by 
\( l=\sqrt{(2-4)^{2}+(3-1)^{2}}=\sqrt{(-2)^{2}+(2)^{2}}\)
\( =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)

(ii) Distance between (-S,7) and (-1,3) is given by
\( l=\sqrt{(-5-(-1))^{2}+(7-3)^{2}}=\sqrt{(-4)^{2}+(4)^{2}}\)
\( =\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) 

(iii) Distance between (a, b) and (-a -b) is given by 
\( l=\sqrt{(a-(-a))^{2}+(b-(-b))^{2}}\)
\( =\sqrt{\left((2 a)^{2}+(2 b)^{2}\right.}=\sqrt{4 a^{2}+4 b^{2}}=2 \sqrt{a^{2}+b^{2}}\) 

Ans : Distance between points (0.0) and (36,15) 
\( =\sqrt{(36-0)^{2}+(15-0)^{2}}=\sqrt{36^{2}+15^{2}}\)
\( =\sqrt{1296+225}=\sqrt{1521}=39\)
Yes, we can find the distance between the given towns A and B Assume town A at origin point (O, O), 
Therefore, town a Rill be at point (36, 15) with respect to town A. 
And hence, as calculated above, the distance between town A and B Bill be 39 km 

Ans : Let the points (1, 5), (2, 3), and (-2, -11) be representing the vertices A, B, and C of the given triangle respectively. 
Let A=(1,5) , B =(2,3) , C=(-2,-11)
\( \therefore A B=\sqrt{(1-2)^{2}+(5-3)^{2}}=\sqrt{5}\)
\(BC=\sqrt{(2-(-2))^{2}+(3-(-11))^{2}}=\sqrt{4^{2}+14^{2}}=\sqrt{16+196}=\sqrt{212}\)
\( CA=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}}=\sqrt{3^{2}+16^{2}}=\sqrt{9+256}=\sqrt{265}\)
Since \( \mathrm{AB}+\mathrm{BC} \neq \mathrm{CA}\)
Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear, 

Ans : Let the points (5, -2), (6, 4), and (7, -2) are representing the vertices A, B, and C  of the given triangle respectively. 
\( A B=\sqrt{(5-6)^{2}+(-2-4)^{2}}=\sqrt{(-1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}\)
\( B C=\sqrt{(6-7)^{2}+(4-(-2))^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{1+36}=\sqrt{37}\)
\( \mathrm{CA}=\sqrt{(5-7)^{2}+(-2-(-2))^{2}}=\sqrt{(-2)^{2}+0^{2}}=2\)
Therefore  AB=BC
As two sides are equal in length, therefore, ABC is an isosceles triangle. 

Ans : \( A B=\sqrt{(3-6)^{2}+(4-7)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
\( B C=\sqrt{(6-9)^{2}+(7-4)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
\( C B=\sqrt{(9-6)^{2}+(4-1)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
\( A D=\sqrt{(3-6)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
\( A C=\sqrt{(3-9)^{2}+(4-4)^{2}}=\sqrt{(-6)^{2}+0^{2}}=6\)
\( \mathrm{BD}=\sqrt{(6-6)^{2}+(7-1)^{2}}=\sqrt{0^{2}+(6)^{2}}=6\)

It can be observed that all sides of this quadrilateral ABCD are of the same length 
and also the diagonals are Of the sarne length, 
Therefore, ABCD is a square and hence, Champa correct 

Ans : (i) Let one points (-1, -2), (1, 0), (-1, 2), and (-3, O) be representing the vertices 
A, B, C, and D Of the given quadrilateral respectively, 
\( \therefore AB=\sqrt{(-1-1)^{2}+(-2-0)^{2}}=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)
\( B C=\sqrt{(1-(-1))^{2}+(0-2)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)
\( C D=\sqrt{(-1-(-3))^{2}+(2-0)^{2}}=\sqrt{(2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)
\( A D=\sqrt{(-1-(-3))^{2}+(-2-0)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)
Diagonal \( A C=\sqrt{(-1-(-1))^{2}+(-2-2)^{2}}=\sqrt{0^{2}+(-4)^{2}}=\sqrt{16}=4\)
Diagonal \( =\sqrt{(1-(-3))^{2}+(0-0)^{2}}=\sqrt{(4)^{2}+0^{2}}=\sqrt{16}=4\)
It can be observed that all sides of this quadrilateral are of the same length end also, 
the diagonals are of the same length. Therefore, the given points are the vertices of 
a square.

(ii)Let the points(- 3, 5), (3, 1), (O, 3), and (-1,-4) be representing the vertices A, B, C, and D Of the given quadrilateral respectively. 
\( A B=\sqrt{(-3-3)^{2}+(5-1)^{2}}=\sqrt{(-6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13}\)
\(\mathrm{BC}=\sqrt{(3-0)^{2}+(1-3)^{2}}=\sqrt{(3)^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}\)
\(\mathrm{CD}=\sqrt{(0-(-1))^{2}+(3-(-4))^{2}}=\sqrt{(1)^{2}+(7)^{2}}=\sqrt{1+49}=\sqrt{50}=5 \sqrt{2}\)
\( A D=\sqrt{(-3-(-1))^{2}+(5-(-4))^{2}}=\sqrt{(-2)^{2}+(9)^{2}}=\sqrt{4+81}=\sqrt{85}\)
It can be observed that all sides of this quadrilateral are of different lengths 
Therefore, it can be said that it is only a general quadrilateral, and not specific such 
as square, rectangle, etc. 

(iii)Let Y-,e points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, a, 
C, and D of the given quadrilateral respectively. 
\(\mathrm{AB}=\sqrt{(4-7)^{2}+(5-6)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9+1}=\sqrt{10}\)
\( B C=\sqrt{(7-4)^{2}+(6-3)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}\)
\(\mathrm{CD}=\sqrt{(4-1)^{2}+(3-2)^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10}\)
\(\mathrm{AD}=\sqrt{(4-1)^{2}+(5-2)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}\)
\( A C=\sqrt{(4-4)^{2}+(5-3)^{2}}=\sqrt{(0)^{2}+(2)^{2}}=\sqrt{0+4}=2\)
\(\mathrm{CD}=\sqrt{(7-1)^{2}+(6-2)^{2}}=\sqrt{(6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52}=13 \sqrt{2}\)
It can be observed that opposite sides of this quadrilateral are of the same length. 
However, the diagonals are Of different lengths, Therefore, the given points are the 
vertices Of a parallelogram 

Ans : We have to find a point on x-axis. Therefore y-coordinate will be 0. 
Let the point on x-axis be (x,0)
Distance between (x,0) , (2,-5)=\( \sqrt{(x-2)^{2}+(0-(-5))^{2}}=\sqrt{(x-2)^{2}+(5)^{2}}\)
Distance between (x,0) , (-2,-9)= \( \sqrt{(x-(-2))^{2}+(0-(-9))^{2}}=\sqrt{(x+2)^{2}+(9)^{2}}\)
By the given condition, these distances are equal in measure. 
\( \sqrt{(x-2)^{2}+(5)^{2}}=\sqrt{(x+2)^{2}+(9)^{2}}\)
\( (x-2)^{2}+25=(x+2)^{2}+81\)
\( x^{2}+4-4 x+25=x^{2}+4+4 x+81\)
8x=25-81
8x=-56
x=-7
Therefore the point is (-7,0) 

Ans : It is given that the distance between (2 -3) and (10, y) is 10. 
Therefore \( \sqrt{(2-10)^{2}+(-3-y)^{2}}=10\)
\( \sqrt{(-8)^{2}+(3+y)^{2}}=10\)
\( 64+(y+3)^{2}=100\)
\( (y+3)^{2}=36\)
\( y+3=\pm 6\)
\( y+3=6 \text { or } y+3=-6\)
Therefore ,y=3 or -9 

Ans : PQ= QR
\( \sqrt{(5-0)^{2}+(-3-1)^{2}}=\sqrt{(0-x)^{2}+(1-6)^{2}}\)
\( \sqrt{(5)^{2}+(-4)^{2}}=\sqrt{(-x)^{2}+(-5)^{2}}\)
\( \sqrt{25+16}=\sqrt{x^{2}+25}\)
\( 41=x^{2}+25\)
\( 16=x^{2}\)
\( x=\pm 4\)
Therefore , point R is (4,6) or (-4,6).
When the point is ( 4,6 )
PQ=\( \sqrt{(5-4)^{2}+(-3-6)^{2}}=\sqrt{1^{2}+(-9)^{2}}=\sqrt{1+81}=\sqrt{82}\)
qR = \( \sqrt{(0-4)^{2}+(1-6)^{2}}=\sqrt{(-4)^{2}+(-5)^{2}}=\sqrt{16+25}=\sqrt{41}\)
When the point R is (-4,6)
PR=\( \sqrt{(5-(-4))^{2}+(-3-6)^{2}}=\sqrt{(9)^{2}+(-9)^{2}}=\sqrt{81+81}=9 \sqrt{2}\)
qR= \( \sqrt{(0-(-4))^{2}+(1-6)^{2}}=\sqrt{(4)^{2}+(-5)^{2}}=\sqrt{16+25}=\sqrt{41}\) 

Ans : Point (x, y) is equidistant from (3, 6) and (-3, 4). 
\( \therefore \sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x-(-3))^{2}+(y-4)^{2}}\)
\( \sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}}\)
\( (x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}\)
\( x^{2}+9-6 x+y^{2}+36-12 y=x^{2}+9+6 x+y^{2}+16-8 y\)
\( 36-16=6 x+6 x+12 y-8 y\)
\( 20=12 x+4 y\)
\( 3 x+y=5\)
\( 3 x+y-5=0\) 

Ans : Let P(x, y) be the required point. using the section formula, we obtain 
\( x=\frac{2 \times 4+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)
\( y=\frac{2 \times(-3)+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Therefore, the point is (1, 3). 

Ans : 

Let P\( \left(x_{1}, y_{1}\right)\) and Q\(\left(x_{2}, y_{2}\right)\) are the points of trisection of the line segment joining 
the given points i.e., AP PQ Q3 
Therefore, point p divides AB internally in the ratio 1:2. 
\( x_{1}=\frac{1 \times(-2)+2 \times 4}{1+2}\)
\( y_{2}=\frac{2 \times(-3)+1 \times(-1)}{2+1}\)
\( x_{2}=\frac{-4+4}{3}=0\)
\( y_{2}=\frac{-6-1}{3}=\frac{-7}{3}\)
\( \mathrm{Q}\left(x_{2}, y_{2}\right)=\left(0,-\frac{7}{3}\right)\) 

Ans : It can be observed that Niharika posted the green flag at \( \frac{1}{4}\) of the distance AD i.e., \( \left(\frac{1}{4} \times 100\right) m=25\)
Similarly, Preet posted red flag at  \( \frac{1}{5}\) of the distance i.e. \( \left(\frac{1}{5} \times 100\right) m=20\) metre from the starting point of 8th line. Therefore , the coordinate of this point R are (8,20).
Distance between these flags by using distance formula = GR
\( =\sqrt{(8-2)^{2}+(25-20)^{2}}=\sqrt{36+25}=\sqrt{61} \mathrm{m}\)
The point at which Rashmi should post her bue flag is the mid-point of the line 
joining these points. Let this point be A (x, y). 
\( x=\frac{2+8}{2}\) , \( y=\frac{25+20}{2}\)
\( x=\frac{10^{2}}{2}=5\) , \( y=\frac{45}{2}=22.5\)
Hence \( A(x, y)=(5,22.5)\)
Therefore , rashmi should post her blue flag at 22.5m on 5th line. 

Ans : Let the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by 
point (-1, 6) be k : 1. 
Therefore  \( -1=\frac{6 k-3}{k+1}\)
\( -k-1=6 k-3\)
7k=2
\( k=\frac{2}{7}\)
Therefore , the required ratio is 2:7. 

Ans : Let the ratio in which the line segment joining A (1, -5) and B (-4, 5) is divided by X-axis be k:1.
Therefore , the coordinates of the point of division is \( \left(\frac{-4 k+1}{k+1}, \frac{5 k-5}{k+1}\right)\)
When we know that y -coordinates of any point on x-axis is 0.
\( \therefore \frac{5 k-5}{k+1}=0\)
k=1
Therefore x-axis divides it in the ratio 1:1.
Division point = \( \left(\frac{-4(1)+1}{1+1}, \frac{5(1)-5}{1+1}\right)=\left(\frac{-4+1}{2}, \frac{5-5}{2}\right)=\left(\frac{-3}{2}, 0\right)\) 

Ans : 
Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a 
parallelogram ABCD. 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 
\( \left(\frac{1+x}{2}, \frac{2+6}{2}\right) \Rightarrow\left(\frac{x+1}{2}, 4\right)\)
If O is the mid-point of BD, then the coordinates of O are 
\( \left(\frac{4+3}{2}, \frac{5+y}{2}\right) \Rightarrow\left(\frac{7}{2}, \frac{5+y}{2}\right)\)
Since both the coordinates are of the sane point O, 
\( \therefore \frac{x+1}{2}=\frac{7}{2}\)
And \(9 4=\frac{5+y}{2}\)
x+1=7 and 5+y=8
x=6 and y=3 

Ans : Let the coordinates of point A be (x, y). 
Mid-point of AS is (2, -3), which is the center of the circle, 
\( \therefore(2,-3)=\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\)
\( \Rightarrow \frac{x+1}{2}=2\) and \( \frac{y+4}{2}=-3\)
x+1=4 and y+4=-6
x=3 and y=-10
Therefore , the coordinates are (3 , -10)