Determine whether the given points are collinear.
(1) A0,2), B1,-0.5), C(2,-3)
Answers
Answer:
find distance between AB,BC,CA
IF AB=BC=CA then 3 points are collinear
distance formula√(x2-x1)^2+(y2-y1)^2
Answer:
The slopes of AB and AC should agree:
A(0,2),B(1,−0.5),C(2,−3)A(0,2),B(1,−0.5),C(2,−3)
slope(AB)=−0.5−21−0=−2.5slope(AB)=−0.5−21−0=−2.5
slope(AC)=−3−22−0=−2.5
Those agree, so we have collinearity.
An alternative that’s essentially the same but avoids division is the point-point form of a line through (a,b)(a,b) and (c,d):(c,d):
(c−a)(y−b)=(d−b)(x−a)
We can consider that the relationship between three points on a line, in our case (a,b)=(0,2),(c,d)=(1,−0.5),(x,y)=(2,−3).(a,b)=(0,2),(c,d)=(1,−0.5),(x,y)=(2,−3). Checking.
(c−a)(y−b)=(1−0)(−3−2)=−5(c−a)(y−b)=(1−0)(−3−2)=−5
(d−b)(x−a)=(−0.5−2)(2−0)=−5
Now to the distance formula. For collinear points, two of the distances should add two the third. Looking at the points we check |AB|+|BC|=|AC|.|AB|+|BC|=|AC|. This is in general a pain because of the square roots; let’s try it.
AB2=(1−0)2+(−0.5−2)2=1+6.25=7.25=29/4AB2=(1−0)2+(−0.5−2)2=1+6.25=7.25=29/4
AC2=(2−0)2+(−3−2)2=29AC2=(2−0)2+(−3−2)2=29
BC2=(2−1)2+(−3−−0.5)2=29/4= 29 root
P+Q±2PQ−−−√=RP+Q±2PQ=R
(R−P−Q)2=4PQ(R−P−Q)2=4PQ
That’s called the Triple Quad Formula and it allows us to check the squared lengths directly for collinearity. The assignment of the squared sides to P,Q,RP,Q,R doesn’t matter; we can choose a way that makes the arithmetic easier, say R=P=29/4,R=P=29/4, Q=29.Q=29.
R−P−Q)2=(29/4−29/4−29)2=292(R−P−Q)2=(29/4−29/4−29)2=292
4PQ=4(29/4)(29)=292✓4PQ=4(29/4)(29)=292✓
These agree so we have collinearity. With the TQF we don’t have to work with any square roots or even figure out which is the longest side to check collinearity.
A bonus from the TQF is the difference of the two sides is sixteen times the area of the triangle when the points aren’t collinear.
16Δ2=4PQ−(R−P−Q)216Δ2=4PQ−(R−P−Q)2