Please help me find y

Question

Find all y such that the distance between the points (2,−9) and (1,y) is 15.

 

I have no idea how to do this problem...

 

Answer 1

Hi, Cassie. We can use the Distance Formula:

 

D = sqrt [(y₂ - y₁)² + (x₂ - x₁)²]

 

We know that D = the distance, which is 15. So, we plug into the above formula. 

 

15 = sqrt [(y - (-9))² + (1 - 2)²]

15 = sqrt[(y + 9)² + 1] = sqrt (y² + 18y + 81 + 1)

15 = sqrt (y² + 18y + 82)

225 = y² + 18y + 82 if we square both sides of the previous equation

y² + 18y - 143 = 0

This cannot be factored, so you will need to use the Quadratic Formula to solve this equation for y. 

Once you do this, you should get:

 

x = -9 + 4 sqrt (14), -9 - 4 sqrt (14)

Answer 2

Use the distance formula:

 

 √[(1-2)²+(y-(-9))²] = 15

 

 √[1+y²+18y+81] = 15

 

   Square both sides to get y²+18y+82 = 225

 

                                      y²+18y-143=0

 

Now, use the quadratic formula to determine the values of y.  

 

Answer 3

Hi Cassie,

 

All the points in a "plane" that are equidistant from (2, -9) lie on a circle with center at (2,-9). In this case the required distance from the center is 15.

 

The equation of that circle is: (x-2)² + (y+9)² = 225. You want to know what are the y-values when x= 1, i.e. what points of the form (1,y) lie on the circle. Substitute 1 for x in the above equation and solve for y.

 

1 + y² +18y + 81 = 225

 

y² + 18y - 143 = 0.

 

Time to use the quadratic formula!