# Ex.7.1 Q6 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) $$(-1, -2)$$, $$(1, 0)$$, $$(-1, 2)$$, $$(-3, 0)$$

(ii) $$(-3, 5)$$,$$(3, 1)$$, $$(0, 3)$$, $$(-1, -4)$$

(iii) $$(4, 5)$$, $$(7, 6)$$, $$(4, 3)$$, $$(1, 2)$$

## Text Solution

Reasoning:

A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.

What is known?

The $$x$$ and $$y$$ co-ordinates of the points between which the distance is to be measured.

What is Unknown?

To determine the type of quadrilateral formed (if any) by the given co-ordinates .

Steps:

(i) Given,

Let the points $$(-1, -2)$$, $$(1, 0)$$, $$(-1, 2)$$, and $$(-3, 0)$$ represent the vertices $$A$$, $$B$$, $$C$$, and $$D$$ of the given quadrilateral respectively.

We know that the distance between the two points is given by the Distance Formula,

\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad ...\,{\text{Equation}}\,(1)\end{align}

To find $$AB$$ i.e. Distance between Points $$A$$ $$(-1, -2)$$ and $$B$$ $$(1, 0)$$

• $$x_1 = -1$$
• $$y_1 = -2$$
• $$x_2 = 1$$
• $$y_2 = 0$$

By substituting the values in the Equation (1)

\begin{align}∴ \; AB& = \sqrt {{{( - 1 - 1)}^2} + {{( - 2 - 0)}^2}} \\\;\;\;\;\;\;\;\;\, &= \sqrt {{{( - 2)}^2} + {{( - 2)}^2}} \\\;\;\;\;\;\;\;\;\,& = \sqrt {4 + 4} \\\;\;\;\;\;\;\;\;\, &= \sqrt 8 \\\;\;\;\;\;\;\;\;\, &= 2\sqrt 2 \end{align}

To find $$BC$$ i.e. Distance between Points $$B\; (1, 0)$$ and $$C \;(-1, 2)$$

• $$x_1 = 1$$
• $$y_1 = 0$$
• $$x_2 = -1$$
• $$y_2 = 2$$

By substituting the values in the Equation (1)

\begin{align}BC &= \sqrt {{{(1 - ( - 1))}^2} + {{(0 - 2)}^2}} \\ &= \sqrt {{{(2)}^2} + {{( - 2)}^2}} \\ &= \sqrt {4 + 4} \\& = \sqrt 8 \\&= 2\sqrt 2 \end{align}

To find $$AD$$ i.e. Distance between Points $$D\; (-3, 0)$$ and $$A \;(-1, -2)$$

• $$x_1 = -3$$
• $$y_1 = 0$$
• $$x_2 = -1$$
• $$y_2 = -2$$

By substituting the values in the Equation (1)

\begin{align}AD& = \sqrt {{{( - 1 - ( - 3))}^2} + {{( - 2 - 0)}^2}} \\ &= \sqrt {{{(2)}^2} + {{( - 2)}^2}} \\& = \sqrt {4 + 4} \\& = \sqrt 8 \\&= 2\sqrt 2 \end{align}

To find $$CD$$ i.e. Distance between Points $$C \;(-1, 2)$$ and $$D \;(-3, 0)$$

• $$x_1 = -1$$
• $$y_1 = 2$$
• $$x_2 = -3$$
• $$y_2 = 0$$

By substituting the values in the Equation (1)

\begin{align}CD &= \sqrt {{{( - 1 - ( - 3))}^2} + {{(2 - 0)}^2}} \\& = \sqrt {{{(2)}^2} + {{(2)}^2}} \\& = \sqrt {4 + 4} \\ &= \sqrt 8 \\&= 2\sqrt 2 \end{align}

To find $$AC$$ i.e. Distance between Points $$A \;(-1, -2)$$ and $$C\; (-1, 2)$$

• $$x_1 = -1$$
• $$y_1 = -2$$
• $$x_2 = -1$$
• $$y_2 = 2$$

By substituting the values in the Equation (1), we get

\begin{align}Diagonal\; AC& = \sqrt {{{( - 1 - ( - 1))}^2} + {{( - 2 - 2)}^2}} \\& = \sqrt {{0^2} + {{( - 4)}^2}} \\ &= \sqrt {16} \\&= 4\end{align}

To find $$BD$$ i.e. Distance between Points $$B \;(1, 0)$$and$$D\; (-3, 0)$$

• $$x_1 = 1$$
• $$y_1 = 0$$
• $$x_2 = -3$$
• $$y_2 = 0$$

By substituting the values in the Equation (1)

\begin{align} \text{Diagonal}\; BD &= \sqrt {{{(1 - ( - 3))}^2} + {{(0 - 0)}^2}} \\ &= \sqrt {{{(4)}^2} + {0^2}} \\& = \sqrt {16} \\ &= 4\end{align}

It can be observed that all 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.

(ii) Steps:

Let the points $$(-3, 5)$$, $$(3, 1)$$, $$(0, 3)$$, and $$(-1, -4)$$ represent the vertices $$A$$, $$B$$, $$C$$, and $$D$$ of the given quadrilateral respectively. We know that the distance between the two points is given by the Distance Formula,

\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad ...\,{\text{Equation}}\,(1)\end{align}

To find $$AB$$ i.e. Distance between Points $$A \;(-3, 5)$$ and $$B\; (3, 1)$$

• $$x_1 = -3$$
• $$y_1 = 5$$
• $$x_2 = 3$$
• $$y_2 = 1$$

By substituting the values in the Equation (1)

\begin{align}AB& = \sqrt {{{( - 3 - 3)}^2} + {{(5 - 1)}^2}} \\ &= \sqrt {{{( - 6)}^2} + {{(4)}^2}} \\& = \sqrt {36 + 16} \\ &= \sqrt {52} \\ &= 2\sqrt {13} \end{align}

To find $$BC$$ i.e. Distance between the Points $$B\;(3, 1)$$ and $$C\; (0, 3)$$

• $$x_1 = 3$$
• $$y_1 = 1$$
• $$x_2 = 0$$
• $$y_2 = 3$$

By substituting the values in the Equation (1)

\begin{align}BC& = \sqrt {{{(3 - 0)}^2} + {{(1 - 3)}^2}} \\ &= \sqrt {{{(3)}^2} + {{( - 2)}^2}} \\ &= \sqrt {9 + 4} \\& = \sqrt {13} \end{align}

To find $$CD$$ i.e. Distance between Points $$C \; (0, 3)$$ and $$D \;(-1, -4)$$

• $$x_1 = 0$$
• $$y_1 = 3$$
• $$x_2 = -1$$
• $$y_2 = -4$$

By substituting the values in the Equation (1)

\begin{align}CD& = \sqrt {{{(0 - ( - 1))}^2} + {{(3 - ( - 4))}^2}} \\& = \sqrt {{{(1)}^2} + {{(7)}^2}} \\ &= \sqrt {1 + 49} \\ &= \sqrt {50} \\ &= 5\sqrt 2 \end{align}

To find $$AD$$ i.e. Distance between Points $$A \;(-3, 5)$$ and $$B\; (-1, -4)$$

• $$x_1 = -3$$
• $$y_1 = 5$$
• $$x_2 = -1$$
• $$y_2 = -4$$

By substituting the values in the Equation (1)

\begin{align}AD &= \sqrt {{{( - 3 - ( - 1))}^2} + {{(5 - ( - 4))}^2}} \\& = \sqrt {{{( - 2)}^2} + {{(9)}^2}} \\ &= \sqrt {4 + 81} \\ &= \sqrt {85} \end{align}

\begin{align}AB \ne BC \ne AC \ne AD\end{align}

Also, by plotting the graph it looks like as below: By the graph above,

$$A$$, $$B$$, $$C$$ are collinear, So, no quadrilateral can be formed from these points

(iii) Steps:

• Let the points $$(4, 5)$$, $$(7, 6)$$, $$(4, 3)$$, and $$(1, 2)$$ be representing the vertices $$A$$, $$B$$, $$C$$, and $$D$$ of the given quadrilateral respectively.

We know that the distance between the two points is given by the Distance Formula,

\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad ...\;{\text{Equation}}\;(1)\end{align}

To find $$AB$$ i.e. Distance between Points $$A \;(4, 5)$$ and $$B\; (7, 6)$$

• $$x_1 = 4$$
• $$y_1 = 5$$
• $$x_2 = 7$$
• $$y_2 = 6$$

By substituting the values in the Equation (1)

\begin{align}AB &= \sqrt {{{(4 - 7)}^2} + {{(5 - 6)}^2}} \\ &= \sqrt {{{( - 3)}^2} + {{( - 1)}^2}} \\& = \sqrt {9 + 1} \\ &= \sqrt {10} \end{align}

To find $$BC$$ i.e. Distance between Points $$B \;(7, 6)$$ and $$C\; (4, 3)$$

• $$x_1 = 7$$
• $$y_1 = 6$$
• $$x_2 = 4$$
• $$y_2 = 3$$

By substituting the values in the Equation (1)

\begin{align}BC &= \sqrt {{{(7 - 4)}^2} + {{(6 - 3)}^2}} \\& = \sqrt {{{(3)}^2} + {{(3)}^2}} \\& = \sqrt {9 + 9} \\ &= \sqrt {18} \end{align}

To find $$CD$$ i.e. Distance between Points $$C\; (4, 3)$$ and $$D \;(1, 2)$$

• $$x_1 = 4$$
• $$y_1 = 3$$
• $$x_2 = 1$$
• $$y_2 = 2$$

By substituting the values in the Equation (1)

\begin{align}CD &= \sqrt {{{(4 - 1)}^2} + {{(3 - 2)}^2}} \\& = \sqrt {{{(3)}^2} + {{(1)}^2}} \\& = \sqrt {9 + 1} \\& = \sqrt {10} \end{align}

To find $$AD$$ i.e. Distance between Points $$A \;(4, 5)$$ and $$D \;(1, 2)$$

• $$x_1 = 4$$
• $$y_1 = 5$$
• $$x_2 = 1$$
• $$y_2 = 2$$

By substituting the values in the Equation (1)

\begin{align}AD &= \sqrt {{{(4 - 1)}^2} + {{(5 - 2)}^2}} \\ &= \sqrt {{{(3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\& = \sqrt {18} \end{align}

To find $$AC$$ i.e. Distance between Points $$A \;(4, 5)$$ and $$C \;(4, 3)$$

• $$x_1 = 4$$
• $$y_1 = 5$$
• $$x_2 = 4$$
• $$y_2 = 3$$

By substituting the values in the Equation (1)

\begin{align}\text{Diagonal} \;AC &= \sqrt {{{(4 - 4)}^2} + {{(5 - 3)}^2}} \\ &= \sqrt {{{(0)}^2} + {{(2)}^2}} \\ &= \sqrt {0 + 4} \\ &= 2\end{align}

To find $$BD$$ i.e. Distance between Points $$B \;(7, 6)$$ and $$D \;(1, 2)$$

• $$x_1 = 7$$
• $$y_1 = 6$$
• $$x_2 = 1$$
• $$y_2 = 2$$

By substituting the values in the Equation (1)

\begin{align}\text{Diagonal}\; BD& = \sqrt {{{(7 - 1)}^2} + {{(6 - 2)}^2}} \\& = \sqrt {{{(6)}^2} + {{(4)}^2}} \\ &= \sqrt {36 + 16} \\ &= \sqrt {52} \\ &= 13\sqrt 2 \end{align}

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.

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