Question

Find x such that the line segment determined by (x, -7) and (-2, -19) is 15 units long.

Answer 1

Suppose we have two points (a, b) and (c, d).
The distance between the two points will be given by sqrt((d - b)^2 + (c - a)^2)

Thus, distance between (x, -7) and (-2, -19) will be:

sqrt((-7 - (-19))^2 + (x - (-2))^2)
= sqrt((-7 + 19)^2 + (x + 2)^2)
= sqrt(12^2 + (x + 2)^2)
= sqrt(144 + (x + 2)^2)

Since the distance is given as 15 units long, we have:

sqrt(144 + (x + 2)^2) = 15
144 + (x + 2)^2 = 15^2
144 + (x + 2)^2 = 225
(x + 2)^2 = 225 - 144
(x + 2)^2 = 81
x + 2 = sqrt(81)
x + 2 = 9 or x + 2 = -9
x = 9 - 2 or x = -9 - 2
x = 7 or x = -11

Hence, the two possible values of x are 7 and -11.

Answer 2

delta y=19-7=12

distans=deltax^2 + deltay^2

15^2=12^2+sumthun=225-144=81

deltax=9  (3-4-5) triangel