Question

A(5, 1) B(x,7) C (8, 2) are three points such that AB=2AC find all possible values of x​

Answer 1

x = 1

Step-by-step explanation:

Distance between A and C =√((x2)2-(x1)2)2 +((y2)2-(y1)2)2

= √((8)2-(5)2) + ((2)2-(1)2)

√(64-25)+(4-1)

√39+3

√42=6.48(approx.)= 6

B: x = 1

y = 7

distance between A and B = 5 + 7 = 12

Therefore AB = 2AC as AB = 12 and AC = 6