Confusing directions?

• Since the points are variable, we can't plot them. Also taking examples wouldn't be enough for a confirmed answer for all cases.

• We know that three non-collinear points make a triangle.

• Checking for non-collinearity...
• slope of PQ =
• $\frac{a-(a+c)}{b-(a-b)}$ = $\frac{-c}{2b-a}$
• and
• slope of QR =
• $\frac{(a+b)-a}{(b+c)-b}$ = $\frac{b}{c}$
• To be collinear their slopes must be equal.

• Hence $\frac{-c}{2b-a}$ = $\frac{b}{c}$

• or $-c^2=2b^2-ab$

• or $c^2-ab= -2b^2$
• but $c^2>ab$ , hence $c^2-ab > 0$
• but $-2b^2$ is negative.
• also $c^2-ab=-2b^2=0$ is not possible as that would imply $b=0$ and $c=0$ which implies Q and R are same points.

• Hence their slopes can't be equal.

• Therefore these three points are always non-collinear, and hence form a triangle.

• We know area of a triangle given the coordinates of its three vertices, as the determinant of

multiplied by 1/2.

• This is derived by taking the vertices in anticlockwise direction.
• But if we take the points in clockwise direction, two points would be interchanged, and the determinant would be

=

multiplied with -1.

• Hence if we calculate the area of triangle using this determinant in the order PQR and it turns out to be positive, it means PQR is in anticlockwise direction, else it is in clockwise direction.

• For this quesion, area of triangle

• PQR = $\frac{1}{2}$ * { (a-b)(-b)+b(b-c)+(b+c)(c) }
• = $\frac{1}{2}$ * {-ab+b^2+b^2-bc+bc+c^2 }
• = $\frac{1}{2}$ * { 2b^2+c^2-ab}
• since $c^2-ab>0$ and $2b^2$ is also positive
• area of triangle PQR is positive.

• Hence direction of travel from PQR would be anticlockwise irrespective of values of a,b and c.

• Any disagreements are mostly welcome.