The points A (3, 0), B (4, 5) and C (6, 7) are collinear.

1.	The points A (3, 0), B (4, 5) and C (6, 7) are collinear.
A.	True
B.	False
Answer» C.

Discussion

If the points A (a, 0), B (0, b) and C (c, 0) are collinear then a = -c.
If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC, then its area is given by \(\frac {1}{2}\) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }
Three points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear only when \(\frac {1}{2}\) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) } = 0.
The points A (3, 0), B (4, 5) and C (6, 7) are collinear.
What will be the value of p, for which the points (-3, p), (-1, 2) and (1, 1) are collinear?
What will be the area of triangle formed by joining the points P (2, 3), Q (4, 6) and R (6, 9)?
What will be the area of triangle PQR formed by joining the midpoints of the sides of the triangle whose vertices are (3, 3), (9, 9) and (4, 6)?
What will be the area of the quadrilateral ABCD whose vertices are A (8, 6), B (9, 0), C (1, 2) and D (3, 4)?
What will be the area of a triangle whose vertices are (3, 1), (0, 4) and (5, 9)?