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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?

User Daree
by
4.8k points

2 Answers

6 votes

Answer:

(-1, 5)

(-1, -3)

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 = √((x_2-x_1)^2 + (y_2-y_1)^2)


5 = √((-1-2)^2 + (a-1)^2)

Solving further, let's square both sides,


5^2 = [√((-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


+/- √(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)

User Temple
by
5.3k points
4 votes

Answer:

A = (-1,5) or (-1,-3)

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)

User Eranga Dissanayaka
by
5.1k points