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

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## Question

Find the distance between the following pairs of points:

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

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

(iii) $$(a, b), (-a, -b)$$

Video Solution
Coordinate Geometry
Ex 7.1 | Question 1

## 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}

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