# Ex.7.1 Q1 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

Find the distance between the following pairs of points:

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

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

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

## Text 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

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