Why do we have to times 2^2 to 13 in the sq root ?

And I don't rlly understand the last part so could u explain it to me? Thank youu

So in the step with 2^2 in the sq root, I’m moving the 2 in 2 √13 so that it is inside the √. Another way to look at it is 2 √13 = 2 * √13 = √(2^2) * √13. From this, we get 2 √13 = √52. From there, I removed the √ from both sides of the equation which allows us to more easily solve for a.

After we find a, we can use the equation for XY to solve for b. This gives us the coordinates for X. After that, we know from the question that X is the same distance from P as Y is from P. So in the last step: P is 6 units away from X on the x-axis (since the x-coordinate of X is 7 while that for P is 1) and P is 4 units away from X on the y-axis (since the y-coordinate of X is 5 while that for P is 1). So Y is the same distance away from P. That’s why in the last step I took the coordinates of P (1,1) and subtracted 6 and 4 respectively to get (-5, -3) as the coordinates of Y.

I hope that clarifies! Please let me know if anything is unclear. :)

Is it 1 must subtract 6 and 4 because it has to be the negative value since X is positive ?

Yep!