Question

Find a so that the two points are 17 units aparts

(-5,8) and (3,a)

please tell me how you found it, i need the help!

Answer 1

Answer: 23 or -7

Explanation:

The distance formula is:

distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Plugging in the values for x1, y1, x2, and y2, we get:

distance = sqrt((3 - (-5))^2 + (a - 8)^2)

Simplifying the expression, we get:

distance = sqrt((8)^2 + (a - 8)^2)

distance = sqrt(64 + (a - 8)^2)

Since the distance between the two points is 17 units, we can set the expression equal to 17 and solve for "a":

sqrt(64 + (a - 8)^2) = 17

64 + (a - 8)^2 = 289

(a - 8)^2 = 225

a - 8 = 15 or a - 8 = -15

a = 23 or a = -7

Answer 2

Therefore, the value of "a" could be 23 or -7 in order for the two points (-5,8) and (3,a) to be 17 units apart.

Explanation:

To find the value of "a" such that the two points (-5,8) and (3,a) are 17 units apart, we can use the distance formula to calculate the distance between the two points.

The distance formula is:

distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Plugging in the values for x1, y1, x2, and y2, we get:

distance = sqrt((3 - (-5))^2 + (a - 8)^2)

Simplifying the expression, we get:

distance = sqrt((8)^2 + (a - 8)^2)

distance = sqrt(64 + (a - 8)^2)

Since the distance between the two points is 17 units, we can set the expression equal to 17 and solve for "a":

sqrt(64 + (a - 8)^2) = 17

64 + (a - 8)^2 = 289

(a - 8)^2 = 225

a - 8 = 15 or a - 8 = -15

a = 23 or a = -7