Step-by-step explanation:
Hey, there!!!
Your question is about showing the points A(1,-1), B(5,2) and C(9,5) as a collinear point.
We generally slope to find weather the points are collinear or not.
So, let's find slope for AB.
[tex]slope(m) = \frac{y2 - y1}{x2 - x1} [/tex]
[tex] slope = \frac{2 + 1}{5 - 1} [/tex]
Therefore, slope of AB = 3/4.
Now, slope of BC.
[tex]slope(m) = \frac{y2 - y1}{x2 - x1} [/tex]
[tex]slope \: (m) = \frac{5 - 2}{9 - 5} [/tex]
Therefore, the slope is 3/4.
now, lastly slope of AC.
[tex]slope(m) = \frac{y2 - y1}{x2 - x1} [/tex]
[tex]slope(m) = \frac{5 + 1}{9 - 1} [/tex]
Therefore, the slope of AC is 3/4.
As all point have same slope. They are collinear point.
Hope it helps...
