plsss can you just try to anwer this, our whole class need it. no one in our class can answer it, it is a 2 points added in my grade if ianswer it :)

Question

a point P(x.y) has a distance 5√2 units from Q (4,-7) and a distance √106 units from R(-6,5), find P?

Answer 1

distance formula = sqrt[(x2-x1)² + (y2-y1)²]  

 

5√2 = √[(4-x1)² + (-7-y1)² ]  --> square both sides

50 = (4-x1)² + (-7 - y1)² --> rearrange to get  50 = (-x + 4)² + (-y - 7)²   --> 

 

repeat steps with second point

106 = (-6 - x)² + (5 - y)²

We have two equations with two unknowns, so:

This process is long and annoying, but I found an answer that satisfied both equations by simply trial and error.

(If you'd like, you can solve this by taking one of the equations, and solving for x or y, and then plugging that into the second equation. )

 If you plug in the below answer you will see it satisfies both:

The answer is (3,0)

 

50 = (-3 + 4)² + (0 - 7)² = 1 + 49 = 50

106 = (-6-3)² + (5)² = 81 + 25 = 106