Question

Please please help so I can finish this

Answer 1

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

Step-by-step 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 $RS = \sqrt{a^{2}+b^{2}+c^{2}-2ac}$.