Question

find the distance between origin (a) 6,-8 (b) 4,-3 (c) 5,-5 (d) 12,-5​

Answer 1

Answer:

Step-by-step explanation:

To Find :-

Distance between origin to :-

a) ( 6 , -8)

b) (4 , -3)

c) (5 , -5)

d) (12 , -5)

Solution :-

Formula Required :-

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

We know that :-

Co-ordinates of origin(O) = (0,0)

a) ( 6 , -8)

x_1 = 0 , y_1 = 0

x_2 = 6 , y_2 = -8

$d = \sqrt{(6-0)^2+(-8-0)^2}$

$d=\sqrt{6^2+(-8)^2$

$d=\sqrt{36+64}$

$d=\sqrt{100}$

$d=10units$

Distance between origin and (6,-8) = 10units.

b) (4 , -3)

x_1 = 0 , y_1 = 0

x_2 = 4 , y_2 = -3

$d=\sqrt{(4-0)^2+(-3-0)^2}$

$d=\sqrt{4^2+(-3)^2}$

$d=\sqrt{16+9}$

$d=\sqrt{25}$

$d=5units$

Distance between origin and (3,-4) = 5units.

c) (5,-5)

x_1 = 0 , y_1 = 0

x_2 = 5 , y_2 = -5

$d=\sqrt{(5-0)^2+(-5-0)^2}$

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

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}units$

Distance between origin and (5,-5) = $5\sqrt{2}units$

d) (12 , -5)

x_1 = 0 , y_1 = 0

x_2 = 12 , y_2 = -5

$d=\sqrt{(12-0)^2+(-5-0)^2}$

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

$d=\sqrt{144+25}$

$d=\sqrt{169}\\\\= 13units$

Distance between origin and (12,-5) = 13units