Question

Find the distance between the following pair of points:

(i) (-6,7) and (-1,-5)

(ii) (a+b,b+c) and (a-b,c-b)

(iii) $(asin\alpha ,-bcos\alpha )and(-acos\alpha ,bsin\alpha )$

(iv) (a,0) and (0,b)

Answer 1

Distance between two points (x_1,y_1)~and~(x_2,y_2)
D = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

(i)

(-6,7) and (-1,-5)
D = $\sqrt{(-1-(-6){)}^2+(-5-7{)}^2}$
=$\sqrt{(5{)}^2+(-12{)}^2}$
=$\sqrt{25+144}$
=$\sqrt{169}$
=$13$

(ii)

(a+b,b+c) and (a-b,c-b)
D = $\sqrt{\left[\right(a-b)-(a+b){]}^2+[(c-b)-(b+c){]}^2}$

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

=$\sqrt{[-2b{]}^2+[-2b{]}^2}$
=$\sqrt{4b^2+4b^2)}$
=$\sqrt{8b^2}$
=$2\sqrt{2}{b}^2$

(iii)

(a sin α,−b cos α) and (−a cos α,b sin α)

D = $\sqrt{(-acos\alpha -asin\alpha {)}^2+(bsin\alpha -(-bcos\alpha ){)}^2}$

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

(iv)

(a,0) and (0,b)
D = $\sqrt{(0-a{)}^2+(b-0{)}^2}$
= $\sqrt{a^2+b^2}$