Question

Please please help so I can finish this

Answer 1

`RS = \sqrt{(b-0)^(2)+(c-a)^(2)}` and RS is simplified to
`RS = \sqrt{a^(2)+b^(2)+c^(2)-2ac}`

Explanation:

We know that , Distance between any two points
`(x_1 , y_1)` and
`(x_2, y_2)` is given by formula :

`Distance = \sqrt{(x_1-x_2)^(2)+(y_1-y_2)^(2)}`

In this question , we have
`R(0,a)` and
`S(b,c)` .

We need to simplify RS which is given by:

`Distance = \sqrt{(x_1-x_2)^(2)+(y_1-y_2)^(2)}`

⇒
`Distance = \sqrt{(x_1-x_2)^(2)+(y_1-y_2)^(2)}`

⇒
`RS = \sqrt{(x_1-x_2)^(2)+(y_1-y_2)^(2)}`

⇒
`RS = \sqrt{(0-b)^(2)+(a-c)^(2)}`

⇒
`RS = \sqrt{(b-0)^(2)+(c-a)^(2)}`

We know that ,
`(a-b)^(2) = a^(2)+b^(2)-2ab`

⇒
`RS = \sqrt{b^(2)+c^(2)+a^(2)-2ac}`

⇒
`RS = \sqrt{a^(2)+b^(2)+c^(2)-2ac}`

∴
`RS = \sqrt{(b-0)^(2)+(c-a)^(2)}` and RS is simplified to
`RS = \sqrt{a^(2)+b^(2)+c^(2)-2ac}`.