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