Find the distance between the following pairs of points
(i) $(2, 3), (4, 1)$
(ii) $(-5, 7), (-1, 3)$
(iii) $(a, b), (-a, -b)$

AcademicMathematicsNCERTClass 10

To do:

We have to find the distance between the given pairs of points.

Solution:

We know that,

The distance between two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Therefore,

(i) The distance between the points $(2, 3)$ and $(4,1)$

$=\sqrt{[4-(2)]^{2}+(1-3)^{2}}$

$=\sqrt{(4-2)^{2}+(-2)^{2}}$

$=\sqrt{(2)^{2}+(-2)^{2}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt2$

The distance between the given pair of points is $2\sqrt2$ units.

(ii) The distance between the points $(-5, 7)$ and $(-1, 3)$

$=\sqrt{[-1-(-5)]^{2}+(3-7)^{2}}$

$=\sqrt{(-1+5)^{2}+(-4)^{2}}$

$=\sqrt{(4)^{2}+(-4)^{2}}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt2$

The distance between the given pair of points is $4\sqrt2$ units.

(iii) The distance between the points $(a, b)$ and $(-a, -b)$

$=\sqrt{[-a-(a)]^{2}+(-b-b)^{2}}$

$=\sqrt{(-a-a)^{2}+(-2b)^{2}}$

$=\sqrt{(-2a)^{2}+(-2b)^{2}}$

$=\sqrt{4a^2+4b^2}$

$=2\sqrt{a^2+b^2}$

The distance between the given pair of points is $2\sqrt{a^2+b^2}$ units.

raja
Updated on 10-Oct-2022 13:22:02

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