NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

To Find: Type of quadrilateral formed
Let the points \( (-1,-2),(1,0),(-1,2) \), and \( (-3,0) \) be representing the vertices \( A, B, C \),
and D of the given quadrilateral respectively

he distance formula is an algebraic expression used to determine the distance between two points with the coordinates \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \).
\( D=\sqrt{ \left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)\(\mathrm{AB}=\left[(-1-1)^{2}+(-2-0)^{2}\right]^{\frac{1}{2}}\)
\(=\sqrt{4+4}=\sqrt{8}\)
\(=2 \sqrt{ 2}\)
\(\mathrm{BC}=\sqrt{ \left[(1+1)^{2}+(0-2)^{2}\right]}\)
\(=\sqrt{4+4}=\sqrt{8}\)
\(=2 \sqrt{ 2}\)
\(\mathrm{AD}=\sqrt{ \left[(-1+3)^{2}+(-2-0)^{2}\right]}\)
\(=\sqrt{4+4}=\sqrt{8}\)
\(=2 \sqrt{ } 2\)
Diagonal \( A C=\sqrt{ \left[(-1+1)^{2}+(-2-2)^{2}\right] }\)
\(\sqrt{0+16}\)
\( =4 \)
Diagonal BD \( =\sqrt{ \left[(1+3)^{2}+(0-0)^{2}\right] }\)
\(\sqrt{16+0}\)
\(=4\)
It is clear that all sides of this quadrilateral are of the same length and the diagonals are of the same length. Therefore, the given points are the vertices of a square

To Find: Type of quadrilateral formed
Let the points \( (-3,5),(3,1),(0,3) \), and \( (-1,-4) \) be representing the vertices \( A, B, C \), and \( D \) of the given quadrilateral respectively.

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates ( \( x_{1}, y_{1} \) ) and \( \left(x_{2}, y_{2}\right) \).
\(D=\sqrt{ \left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
\(A B=\sqrt{ \left[(-3-3)^{2}+(5-1)^{2}\right]}\)
\(\sqrt{36+16}=\sqrt{52}\)
\(=2 \sqrt{ 13}\)
\(\mathrm{BC}=\sqrt{ \left[(3-0)^{2}+(1-3)^{2}\right]}\)
\(=\sqrt{9+4}\)
\(=\sqrt{13}\)
\(\mathrm{CD}=\sqrt{ [(0+1)^2+(3+4)^2]}\)
\(=\sqrt{1+49}=\sqrt{50}\)
\(=5 \sqrt{ 2}\)
\(\mathrm{AD}=\sqrt{ \left[(-3+1)^{2}+(5+4)^{2}\right]}\)
\(=\sqrt{4+81}\)
\(=\sqrt{85}\)
We can observe 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.

To Find: Type of quadrilateral formed
Let the points \( (4,5),(7,6),(4,3) \), and \( (1,2) \) be representing the vertices \( \mathrm{A}, \mathrm{B}, \mathrm{C} \), and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates ( \( x_{1}, y_{1} \) ) and \( \left(x_{2}, y_{2}\right) \).
\(D=\sqrt{ \left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
\(\mathrm{AB}=\sqrt{\left[(4-7)^{2}+(5-6)^{2}\right] }\)
\(=\sqrt{9+1}\)
\(=\sqrt{10}\)
\(\mathrm{BC}=\sqrt{ \left[(7-4)^{2}+(6-3)^{2}\right]}\)
\(=\sqrt{9+9}\)
\(=\sqrt{18}\)
\(\mathrm{CD}=\sqrt{ \left[(4-1)^{2}+(3-2)^{2}\right]}\)
\(=\sqrt{9+1}\)
\(=\sqrt{10}\)
\(\mathrm{AD}=\sqrt{\left[(4-1)^{2}+(5-2)^{2}\right] }\)
\(=\sqrt{9+9}\)
\(=\sqrt{18}\)
\(\text { Diagonal }AC =\sqrt{\left[(4-4)^{2}+(5-3)^{2}\right]}\)
\(=\sqrt{0+4}\)
\(=2\)
\(\text { Diagonal }BD =\sqrt{ \left[(7-1)^{2}+(6-2)^{2}\right]}\)
\(=\sqrt{36+16}\)
\(=\sqrt{52}\)
\(=2 \sqrt{ 13}\)
We can observe 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.