To demonstrate that point P is a point of trisection of the line segment joining points A and B, we need to show that point P divides the line segment AB into three equal parts; in other words, P splits AB such that AP is one third of AB, and PB is two thirds of AB.
Let's start by finding the vector AB, which can be found as follows:
AB = B - A
= (-8, 5) - (1, 2)
= (-8 - 1, 5 - 2)
= (-9, 3)
Now, let's find vectors AP and PB to compare their lengths with AB.
Vector AP is from A to P:
AP = P - A
= (-2, 1) - (1, 2)
= (-2 - 1, 1 - 2)
= (-3, -1)
Vector PB is from P to B:
PB = B - P
= (-8, 5) - (-2, 1)
= (-8 + 2, 5 - 1)
= (-6, 4)
Now let's compare AP and PB with AB to see if the proportional lengths hold.
If P is the first point of trisection, then AP should be one third of AB:
AP should be (1/3)AB, to prove this:
AP = (-3, -1)
(1/3)AB = (1/3)(-9, 3)
= (-3, 1)
Upon comparison, AP does not equal (1/3)AB because we have AP = (-3, -1) and (1/3)AB = (-3, 1). Since the y-component of AP is not the same as the y-component of (1/3)AB, point P does not trisect the line segment AB at the correct ratio.
Therefore, point P is not the point of trisection of the line segment AB.
