Question

Point B has coordinates ​(2​,1​). The​ x-coordinate of point A is negative 1. The distance between point A and point B is 5 units. What are the possible coordinates of point​ A?

Answer 1

Answer:

A = (-1,5) or  (-1,-3)

Step-by-step explanation:

A = (-1,y)    B = (2,1)

 

(Distance from A to B) = √[(-1-2)² + (y-1)²] = 5

                                    =√[9 + y² - 2y + 1] = 5

Squaring on both sides

                                     =  y² - 2y + 10 = 25

                                       =y² - 2y -15 = 0

                                      = (y-5)(y+3) = 0

                                         y = 5 or -3

Therefore, A = (-1,5) or  (-1,-3)

Answer 2

Answer:

(-1, 5)

(-1, -3)

Step-by-step explanation:

Given that coordinates of point B = (2,1) this means, x2 = 2 & y2 = 1.

The x coordinate of point A = -1

D = 5 units

Let's use the equation for distance between points, it is expressed as:

$D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$5 = \sqrt{(-1-2)^2 + (a-1)^2}$

Solving further, let's square both sides,

$5^2 = [\sqrt{(-1-2)^2 + (a-1)^2}]^2$

$25 = (-1-2)^2 + (a-1)^2$

$25 = 9 + (a-1)^2$

Solving further, we have:

$25 - 9 = (a-1)^2$

$16 = (a-1)^2$

$+/- \sqrt{16} = a-1$

$+/-4 = a - 1$

For the possible values, we have:

$a = 4+1 = 5$

$a = - 4 + 1 = -3$

Therefore, the possible coordinates of point​ A are:

(-1, 5)

(-1, -3)