Question

Find the distance between the points A and B in the following:
(i) $A(a+b,b-a),B(a-b,a+b)$
(ii) $A(1,-1),B(-\frac{1}{2},\frac{1}{2})$
(iii) $A(8,-2),B(3,-6)$
(iv) $A(a+b\cdot a-b),B(a-b,-a-b)$

Answer 1

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated

using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
(i) Distance between the points A$(a+b,b-a)$ and B $(a-b,a+b)$

$=\sqrt{{(a-b-a-b)}^2+{(a+b-b+a)}^2}=\sqrt{4b^2+4a^2}=2\sqrt{a^2+b^2}$
(ii) Distance between the points A$(1,-1)$ and B $(-\frac{1}{2},\frac{1}{2})$

$=\sqrt{{(-\frac{1}{2}-1)}^2+{(\frac{1}{2}+1)}^2}=\sqrt{\frac{9}{4}+\frac{9}{4}}=\sqrt{\frac{9}{2}}=3\frac{1}{\sqrt{2}}=3\frac{\sqrt{2}}{2}$

(iii) Distance between the points A$(8,-2)$ and B $(3,-6)$

$=\sqrt{{(3-8)}^2+{(-6+2)}^2}=\sqrt{25+16}=\sqrt{41}$
(iv) Distance between the points A$(a+b,a-b)$ and B $(a-b,-a-b)$

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