In the xy-coordinate plane, the distance between point A (7,9) and point B (4,y) is 5. What is one possible value of y?

Question

Please help!

Answer 1

Use the distance formula between the two points:

 

d² = (x_B-x_A)² + (y_B-y_A)²

 

5² = (4-7)² + (y-9)²

 

25 = 9 + (y-9)²

 

16 = (y-9)²

 

√16 = √(y-9)²

 

±4 = y - 9

 

y = 5 and 13

 

 

Answer 2

To solve this problem, you need to have the distance formula.  It is difficult to type out but I will do my best to explain.

 

d=the square root of (x₂-x₁)2+(y₂-y₁)²

 

You need to solve the equation for y by substituting in all of the values that you know.

 

d=5, x₁=7, x₂=4, y₁=9

5=square root (4-7)²+(y-9)²

 

square both sides:

25=(4-7)²+(y-9)²

 

25=9+(y-9)²

16=(y-9)²

+/-4=y-9

y=13 or 5

Answer 3

The distance formula is

 

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

 

We want to solve for y₂ using some algebra.  Therefore

 

d² = (x₂ - x₁)² + (y₂ - y₁)²

 

d² - (x₂ - x₁)² = (y₂ - y₁)²

 

√[d² - (x₂ - x₁)²] = y₂ - y₁

 

√[d² - (x₂ - x₁)²] + y₁  =  y₂

 

Substitute values into this new form of the distance formula.

 

√[5² - (4 - 7)²] + 9 = y₂

 

√[25 - (-3)²] + 9 = y₂

 

√(25 - 9) + 9 = y₂

 

±√(16) + 9 = y₂

 

±4+9 = y₂

 

13 = y₂

 

5 = y₂