Find the distance between the following pair of points:

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

(i) The two given points are (−6, 7) and (−1, −5)

The distance between these two points is

$d=\sqrt{(-6+1)^{2}+(7+5)^{2}}$

$=\sqrt{(-5)^{2}+(12)^{2}}$

$=\sqrt{25+144}$

 

$=\sqrt{169}$

$d=13$

Hence the distance is 13 units.

(ii) The two given points are $(a+b, b+c)$ and $(a-b, c-b)$

The distance between these two points is

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

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

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

$=\sqrt{8 b^{2}}$

$d=2 b \sqrt{2}$

Hence the distance is $2 b \sqrt{2}$ units.

(iii) The two given points are $(a \sin \alpha,-b \cos \alpha)$ and $(-a \cos \alpha, b \sin \alpha)$

The distance between these two points is

$d=\sqrt{(a \sin \alpha+a \cos \alpha)^{2}+(-b \cos \alpha-b \sin \alpha)^{2}}$

$=\sqrt{a^{2}(\sin \alpha+\cos \alpha)^{2}+b^{2}(-1)^{2}(\cos \alpha+\sin \alpha)^{2}}$

$=\sqrt{a^{2}(\sin \alpha+\cos \alpha)^{2}+b^{2}(\sin \alpha+\cos \alpha)^{2}}$

$=\sqrt{\left(a^{2}+b^{2}\right)(\sin \alpha+\cos \alpha)^{2}}$

$d=(\sin \alpha+\cos \alpha) \sqrt{\left(a^{2}+b^{2}\right)}$

Hence the distance is $(\sin \alpha+\cos \alpha) \sqrt{\left(a^{2}+b^{2}\right)}$

(iv) The two given points are (a, 0) and (0, b)

The distance between these two points is

$d=\sqrt{(a-0)^{2}+(0-b)^{2}}$

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

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

Hence the distance is $\sqrt{a^{2}+b^{2}}$.