# NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1

Coordinate Geometry

Exercise 7.1

## Chapter 7 Ex.7.1 Question 1

Find the distance between the following pairs of points:

(i) \((2, 3), (4, 1)\)

(ii) \((-5, 7), (-1, 3)\)

(iii) \((a, b), (-a, -b)\)

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\(\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\)

**What is Known?**

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

**What is Unknown?**

The distance between the given pairs of points is to be measured.

**Steps:**

(i) \((2, 3), (4, 1)\)

Given,

Let the points be \(A(2, 3)\) and \(B(4, 1)\)

Therefore,

- \(x_1 = 2\)
- \(y_1 = 3\)
- \(x_2 = 4\)
- \(y_2 = 1\)

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

Distance Formula

\(\begin{align}= \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} \;\;...(1)\end{align}\)

Therefore, distance between \(A(2, 3)\) and \(B(4, 1)\) is given by

\[\begin{align}d &= \sqrt {{{(2 - 4)}^2} + {{(3 - 1)}^2}} \\ &= \sqrt {{{( - 2)}^2} + {{(2)}^2}} \\ &= \sqrt {4 + 4} \\ &= \sqrt 8 \\ &= 2\sqrt 2 \end{align}\]

(ii) \((-5, 7), (-1, 3)\)

Distance between \(\begin{align}( - 5,\;7)\;{\text{and}}\;( - 1,\;3)\end{align}\) is given by

\[\begin{align}d &= \sqrt {{{( - 5 - ( - 1))}^2} + {{(7 - 3)}^2}} \\ &= \sqrt {{{( - 4)}^2} + {{(4)}^2}} \\ &= \sqrt {16 + 16} \\ &= \sqrt {32} \\ &= 4\sqrt 2 \end{align}\]

(iii) \((a, b), (-a, -b)\)

Distance between \(\begin{align}(a,\;b)( - a,\; - b)\end{align}\) is given by

\[\begin{align}d &= \sqrt {{{(a - ( - a))}^2} + {{(b - ( - b))}^2}} \\ &= \sqrt {{{(2a)}^2} + {{(2b)}^2}} \\& = \sqrt {4{a^2} + 4{b^2}} \\ &= 2\sqrt {{a^2} + {b^2}} \end{align}\]

## Chapter 7 Ex.7.1 Question 2

Find the distance between the points \((0, 0)\) and \((36, 15)\). Can you now find the distance between the two towns \(A\) and \(B\) discussed in Section 7.2?

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\[\begin{align}= \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\]

**What is Known?**

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

**What is Unknown?**

The distance between the towns \(A\) and \(B\).

**Steps:**

Given:

Let the points be \(A(0,0)\) and \(B(36,15)\)

Hence,

- \(x_1 = 0\)
- \(y_1 = 0\)
- \(x_2 = 36\)
- \(y_2 =15\)

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

\[\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} ...\,(1) \\&= \sqrt {{{(36 - 0)}^2} + {{(15 - 0)}^2}} \\

& = \sqrt {{{36}^2} + {{15}^2}} \\& \text{[By substituting in equation (1)}]\\\\

& = \sqrt {1296 + 225} \\

&= \sqrt {1521} \\

&= 39\end{align}\]

Yes, it is possible to find the distance between the given towns \(A\) and \(B\).

The positions of the towns \(A\) & \(B\) are given by \((0, 0)\) and \((36, 15)\), hence, as calculated above, the distance between town \(A\) and \(B\) will be \(39\,\rm{km}\)

## Chapter 7 Ex.7.1 Question 3

Determine if the points \((1, 5)\), \((2, 3)\) and \((-2, -11)\) are collinear.

**Solution**

**Video Solution**

**Reasoning:**

- Three or more points are said to be collinear if they lie on a single straight line.
- The distance between the two points can be measured using the Distance Formula which is given by: \[\begin{align}&\text{Distance Formula } \\&= \sqrt {{{\left( {{{{x}}_{{1}}} - {{{x}}_{{2}}}} \right)}^2} + {{\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)}^2}} .\end{align}\]

**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 if the \(3\) points are collinear.

**Steps:**

Let the points \((1, 5)\), \((2, 3)\), and \((-2, -11)\) be represented by the \(A\), \(B\), and \(C\) of the given triangle respectively.

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

\[\begin{align} = \sqrt {{{\left( {{{{x}}_{{1}}} - {{{x}}_{{2}}}} \right)}^2} + {{\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)}^2}} ...\,(1)\end{align}\]

To find \(AB\) i.e. the Distance between the Points \(A \;(1, 5)\) and \(B\; (2, 3)\)

- \(x_1 = 1\)
- \(y_1 = 5\)
- \(x_2 = 2\)
- \(y_2 = 3\)

\[\begin{align}\therefore \quad {AB} &= \sqrt {{{(1 - 2)}^2} + {{(5 - 3)}^2}} \\&\text{[By substituting in (1)]}\\\\&=\sqrt 5 \end{align}\]

To find \(BC\) Distance between Points \(B \;(2, 3)\) and \(C\; (-2, -11)\)

- \(x_1 = 2\)
- \(y_1 = 3\)
- \(x_2 = -2\)
- \(y_2 = -11\)

\[\begin{align} \therefore{BC} \!&=\! \sqrt{{{(2 \!-\! (\!-2))}^2}\!+\!{{(3\!+\!(-11))}^2}}\\ &= \sqrt {{4^2} + {{14}^2}} \qquad \\&\text{[By substituting in equation(1)]}\\\\ &= \sqrt {16 + 196} \\&= \sqrt {212} \end{align}\]

To find \(AC\) Distance between Points \(A\; (1, 5)\) and \(C \;(-2, -11)\)

- \(x_1 = 1\)
- \(y_1 = 5\)
- \(x_2 = -2\)
- \(y_2 = -11\)

\[\begin{align} \therefore {CA} &= \sqrt {{{(1\!-\!(-2))}^2}\!+\!{{(5\!+\!(-11))}^2}}\\&= \sqrt {{3^2} + {{16}^2}} \qquad \qquad \\&\text{ [By substituting in equation (1)]}\\\\&= \sqrt {9 + 256}\\&= \sqrt {265} \end{align}\]

Since \(AB + AC\) \(\not=\) \(BC\) and \(AB\) \(\not=\) \(BC\) \(+ AC\) and \(AC\) \(\not=\) \(BC\) Therefore, the points \((1, 5)\), \((2, 3)\), and \((-2, -11)\) are not collinear.

## Chapter 7 Ex.7.1 Question 4

Check whether \((5, -2)\), \((6, 4)\) and \((7, -2)\) are the vertices of an isosceles triangle.

**Solution**

**Video Solution**

**Reasoning:**

An isosceles triangle is a triangle that has two sides of equal length.

To check whether the given points are vertices of an isosceles triangle, the distance between any of the \(2\) points should be the same for two pairs of points.

**What is Known?**

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

**What is Unknown?**

To check whether the given points are the vertices of an isosceles triangle.

**Steps:**

Let the points \((5, -2)\), \((6, 4)\), and \((7, -2)\) represent the vertices \(A\), \(B\), and \(C\) of the given triangle respectively.

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

Distance Formula

\(\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} \;\;...(1)\end{align}\)

To find \(AB\) i.e. Distance between Points \(A\; (5, -2)\) and \(B \;(6, 4)\)

- \(x_1 = 5\)
- \(y_1 = 2\)
- \(x_2 = 6\)
- \(y_2 = 4\)

By substituting the values in the Equation (1)

\[\begin{align}AB &= \sqrt {{{(5 - 6)}^2} + {{( - 2 - 4)}^2}} \\ &= \sqrt {{{( - 1)}^2} + {{( - 6)}^2}} \\ &= \sqrt {1 + 36} \\ &= \sqrt {37} \end{align}\]

To find \(BC\) Distance between Points \(B \;(6, 4)\) and \(C \;(7, -2)\)

- \(x_1 = 6\)
- \(y_1 = 4\)
- \(x_2 = 7\)
- \(y_2 = -2\)

By substituting the values in the Equation (1)

\[\begin{align}BC &= \sqrt {{{(6 - 7)}^2} + {{(4 - ( - 2))}^2}} \\ &= \sqrt {{{( - 1)}^2} + {{(6)}^2}} \\ &= \sqrt {1 + 36} \\ &= \sqrt {37} \end{align}\]

To find \(AC\) i.e. Distance between Points \(A \;(5, -2)\) and \(C \;(7, -2)\)

- \(x_1 = 5\)
- \(y_1 = -2\)
- \(x_2 = 7\)
- \(y_2 = -2\)

\[\begin{align}CA& = \sqrt {{{(5 - 7)}^2} + {{( - 2 - ( - 2))}^2}} \\ &= \sqrt {{{( - 2)}^2} + {0^2}} \\ &= 2\end{align}\]

From the above values of \(AB\), \(BC\) and \(AC\) we can conclude that \(AB\) \(=\) \(BC\). As, two sides are equal in length, therefore, \(ABC\) is an isosceles triangle.

## Chapter 7 Ex.7.1 Question 5

In a classroom, \(4\) friends are seated at the points \(A\), \(B\), \(C\) and \(D\) as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think \(ABCD\) is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

**Solution**

**Video Solution**

**Reasoning:**

To prove that the points \(A\),\(B\),\(C\) and \(D\) from a square, the length of the four sides should be equal and the length of the two diagonals should be the same.

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points between which the distance is to be measured can be deduced from the diagram.

**What is Unknown?**

To verify whether the positions of the four friends form a square or not.

**Steps:**

Let \(A \;(3, 4)\), \(B\; (6, 7)\), \(C \;(9, 4)\), and \(D \;(6, 1)\) be the positions of \(4\) friends.

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

\[\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \;\;\dots(1)\]

To find \(AB\) i.e. Distance between Points \(A \;(3, 4)\) and \(B\; (6, 7)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 =7\)

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

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

To find \(BC\) i.e. Distance between Points \(B (6, 7)\) and \(C (9, 4)\)

- \(x_1 = 6\)
- \(y_1 = 7\)
- \(x_2 = 9\)
- \(y_2 = 4\)

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

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

To find \(CD\) i.e. Distance between Points \(C \; (9, 4)\) and \(D\; (6, 1)\)

- \(x_1 = 9\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\[\begin{align}CB& = \sqrt {{{(9 - 6)}^2} + {{(4 - 1)}^2}} \\ &= \sqrt {{{(3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\ &= \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(AD\) i.e. Distance between Points \(B\; (3, 4)\) and \(D \;(6, 1)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\[\begin{align}AD &= \sqrt {{{(3 - 6)}^2} + {{(4 - 1)}^2}} \\ &= \sqrt {{{( - 3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\&= \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(AC\) i.e. Distance between Points \(A \;(3, 4)\) and \(C\; (9, 4)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 9\)
- \(y_2 = 4\)

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

\(\begin{align}{\text{Diagonal}}\,AC &\!=\! \sqrt {{{(3\! - \!9)}^2} \!+ \!{{(4\! - \!4)}^2}} \\ &= \sqrt {{{( - 6)}^2} + {0^2}} \\ &= 6\end{align}\)

To find \(BD\) Distance between Points \(B \;(6, 7)\) and \(D \;(6, 1)\)

- \(x_1 = 6\)
- \(y_1 = 7\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\(\begin{align}\text{Diagonal} \,BD &\!=\! \sqrt {{{\!(6 \!- \!6)}^2} \!+\! {{(7\! -\! 1)}^2}} \\ &= \sqrt {{0^2} + {{( - 6)}^2}} \\ &= 6\end{align}\)

The four sides \(AB\), \(BC\), \(CD\), and \(AD\) are of same length and diagonals \(AC\) and \(BD\) are of equal length. Therefore, \(ABCD\) is a square and hence, Champa was correct

## Chapter 7 Ex.7.1 Question 6

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)

\((4, 5)\), \((7, 6)\), \((4, 3)\), \((1, 2)\)

**Solution**

**Video 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,

\[\begin{align}\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}\;\;\dots(1)\end{align}\]

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

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

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,

\[\begin{align}\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \;\;\dots(1)\end{align}\]

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,

\[\begin{align}\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \;\;\dots(1)\end{align}\]

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.

## Chapter 7 Ex.7.1 Question 7

Find the point on the \(x\)-axis which is equidistant from \((2, -5)\) and \((-2, 9)\).

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\[\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\]

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points from which the point on the \(x\)-axis is equidistant.

**What is Unknown?**

The point on the \(x\)-axis which is equidistant from \((2, -5)\) and \((-2, 9)\).

**Steps:**

Given,

Since the point is on \(x\)-axis the co-ordinates are \((x, 0)\).

We have to find a point on \(x\)-axis which is equidistant from \(A\; (2, -5)\) and \(B \;(-2, 9)\).

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

\[\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}\;\;\;(1)\end{align}\]

To find the distance between \(PA\), substitute the values of \(P\; (x, 0)\) and \(A\; (2, -5)\) in Equation (1),

\[\begin{align} &= \sqrt {{{(x - 2)}^2} + {{(0 - ( - 5))}^2}} \\ &= \sqrt {{{(x - 2)}^2} + {{(5)}^2}} \end{align}\]

To find the distance between \(PB\), substitute the values of \(P \;(x, 0)\) and \(B \;(-2, 9)\) in Equation (1),

Distance

\[\begin{align}&=\sqrt {{{(x - ( - 2))}^2} + {{(0 - ( - 9))}^2}} \\& = \sqrt {{{(x + 2)}^2} + {{(9)}^2}} \end{align}\]

By the given condition, these distances are equal in measure.

Hence \(PA = PB\)

\(\begin{align}\!\sqrt {{{(\!x \!- \!2)}^2} \!+\! {{(5)}^2}} &= \!\sqrt {{{(x\! + \!2)}^2} \!+ \!{{(9)}^2}} \end{align}\)

Squaring on both sides

\(\begin{align} {(x - 2)^2} + 25 &= {(x + 2)^2} \!+\! 81 \\ {x^2} + 4 - 4x + 25 &= \!{x^2} \!+ \!4 +\! 4x \!+ \!81 \\ 8x &= 25 - \!81 \\ 8x &= - 56 \\x &= - 7\end{align}\)

Therefore, the point equidistant from the given points on the axis is \((-7, 0)\).

## Chapter 7 Ex.7.1 Question 8

Find the values of \(y\) for which the distance between the points \(P\;(2, -3)\) and \(Q \;(10, y)\) is \(10\) units.

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\(\begin{align} =\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\end{align}\)

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points which is at a distance of \(10\) units.

**What is Unknown?**

The values of \(y\) for which the distance between the points \(P\;(2, -3)\) and \(Q \;(10, y)\) is \(10\) units

**Steps:**

Given,

Distance between points \(A \;(2, -3)\) and \(B \;(10, y)\) is \(10\) units.

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

\[\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}}\right)}^2}}\;\;\dots(1)\end{align}\]

By Substituting the values of points \(A \;(2, -3)\) and \(B \;(10, y)\) in Equation (1)

Therefore,

\(\begin{align}\sqrt {{{(2 - 10)}^2} + {{( - 3 - y)}^2}} &= 10\\\sqrt {{{( - 8)}^2} + {{(3 + {\text{y}})}^2}} &= 10\\{\text{Squaring on both sides}}\\64 + {(y + 3)^2} &= 100\\{(y + 3)^2} &= 36\\y + 3 &= \pm 36\\y + 3 &= \pm 6\\y + 3 &= 6\\{\text{or}}\\{\text{y}} + 3 &= - 6\end{align}\)

Therefore, \(\begin{align}y = 3\quad{\text{or}}\quad -y= 9\end{align}\) are the possible values for \(y\)?

## Chapter 7 Ex.7.1 Question 9

If \(Q \;(0, 1)\) is equidistant from \(P \;(5, -3)\) and \(R\; (x, 6)\), find the values of \(x.\) Also find the distances \(QR\) and \(PR\).

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\[\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\]

**What is the known?**

The \(x\) and \(y\) co-ordinates of the points \(P\), \(Q\) and \(R\) between which the distance is to be measured.

**What is the unknown?**

The value of \(x\) and the distance \(QR\) and \(PR\).

**Steps:**

Given,

Since \(Q\; (0, 1)\) is equidistant from \(P\; (5, -3) \)and \(R \;(x, 6)\),\(\begin{align}PQ = QR\end{align}\)

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

\[\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \;\;\dots(1)\end{align}\]

Hence by applying the distance formula for the \(\begin{align}PQ = QR,\end{align}\)we get

\(\begin{align}\!\sqrt {{{(5\! -\! 0)}^2}\! +\! {{( - 3\! - \!1)}^2}} \!&=\! \sqrt {{{(0 - x)}^2} + {{(1 - 6)}^2}} \\\sqrt {{{(5)}^2} + {{( - 4)}^2}} &=\! \sqrt {{{( - x)}^2} + {{( - 5)}^2}} \end{align}\)

By squaring both the sides,

\[\begin{align}25 + 16 &= {x^2} + 25\\\;\;16 &= {x^2}\\\;\;x &= \pm 4\end{align}\]

Therefore, point \(R\) is \((4, 6)\) or \((-4, 6)\).

**Case (1),**

When point R is \((4, 6)\),

Distance between \(P \;(5, -3)\) and \(R\; (4, 6)\) can be calculated using the Distance Formula as ,

\[\begin{align}PR &= \sqrt {{{(5 - 4)}^2} + {{( - 3 - 6)}^2}} \\ &= \sqrt {{1^2} + {{( - 9)}^2}} \\ &= \sqrt {1 + 81} \\ &= \sqrt {82}\end{align}\]

Distance between \(Q\; (0, 1)\) and \(R\; (4, 6)\) can be calculated using the Distance Formula as ,

\[\begin{align}QR &= \sqrt {{{(0 - 4)}^2} + {{(1 - 6)}^2}} \\ &= \sqrt {{{( - 4)}^2} + {{( - 5)}^2}} \\ &= \sqrt {16 + 25} \\& = \sqrt {41}\end{align}\]

**Case (2),**

When point R is \((-4, 6)\),

Distance between \(P\; (5, -3)\) and \(R\; (-4, 6)\) can be calculated using the Distance Formula as,

\[\begin{align}PR& = \sqrt {{{(5 - ( - 4))}^2} + {{( - 3 - 6)}^2}} \\& = \sqrt {{{(9)}^2} + {{( - 9)}^2}} \\ &= \sqrt {81 + 81} \\& = 9\sqrt 2 \end{align}\]

Distance between \(Q \;(0, 1)\) and \(R\; (-4, 6) \) can be calculated using the Distance Formula as ,

\[\begin{align}QR& = \sqrt {{{(0 - ( - 4))}^2} + {{(1 - 6)}^2}} \\& = \sqrt {{{(4)}^2} + {{( - 5)}^2}} \\ &= \sqrt {16 + 25} \\&= \sqrt {41} \end{align}\]

## Chapter 7 Ex.7.1 Question 10

Find a relation between \(x\) and \(y\) such that the point \((x, y)\) is equidistant from the point \((3, 6)\) and \((-3, 4)\).

**Solution**

**Video Solution**

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\(\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\)

**What is the known?**

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

**What is the unknown?**

The relation between \(x\) and \(y\) such that the point \((x, y)\) is equidistant from the point \((3, 6)\) and \((-3, 4)\)

**Steps:**

Let Point \(P(x, y)\) be equidistant from points \(A \;(3, 6)\) and \(B\; (-3, 4)\).

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

\(\begin{align}\!\sqrt {{{\left( {{x_1}\! - {x_2}} \right)}^2}\! +\! {{\left( {{y_1} - {y_2}}\! \right)}^2}}\;\;\dots(1)\end{align}\)

Since they are equidistant, \(PA = PB\)

Hence by applying the distance formula for \(PA = PB\), we get

\(\begin{align} \sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} &= \sqrt {x - {{( - 3)}^2} + {{(y - 4)}^2}} \\\;\;\,\sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} &= \sqrt {{{(x + 3)}^2} + {{(y - 4)}^2}} \end{align}\)

By Squaring, \(\begin{align}PA^2=PB^2\end{align}\)

\[\begin{align}{(x\! -\! 3)^2} \!+\! {(y \!-\! 6)^2} &\!=\! {(x \!+\! 3)^2}\! +\! {(y \!-\! 4)^2}\\ \begin{bmatrix}{x^2} \!+ \!9\! -\! 6x\! +\!\\ {y^2} \!+\! 36\! -\! 12y \end{bmatrix}&\!=\! \begin{bmatrix}{x^2}\! +\! 9 \!+\! 6x\! +\! \\{y^2}\! +\! 16\! -\! 8y\end{bmatrix}\\6x\! +\! 6x\! + \!12y\! -\! 8 y &\!=\! 36\! -\! 16\\12x \!+\! 4y &\!=\!20\\3x \!+\! y &\!=\! 5\\3x\! +\! y\! -\! 5 &\!=\! 0\end{align}\]

Thus, the relation between \(x\) and \(y\) is given by \(\begin{align}3x+ y-5 = 0\end{align}\)