Question

21] If A (6,-1), B (1, 3) and C (k, 8) are three points on a straight line such that AB = BC, find
the value of 'K.
diameter AB on it. Construct tangents at the points A and B​

Answer 1

Step-by-step explanation:

A=1/2[X1(y2-y3)+x2(y3-y1 )+X3(y1-y2)

0=1/2[6(3-8)+ 1(8-(-1)) +k(-1-3)

0×2/1=6(-5)+1 (9)+k(-4)

0= -30 +9-4k

0= -21-4k

4k = -21

k = -21/4