Use your method (from Problem 6) to check if the points R (-5, -1), B (-2, -5) and C (4, -12) are on the same straight line. Now plot both sets of points and check your answers.

Using distance formula,

$$\begin{aligned} RB &= \sqrt{(-2+5)^2 + (-5+1)^2} \\ &= \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5. \end{aligned}$$

$$\begin{aligned} BC &= \sqrt{(4+2)^2 + (-12+5)^2} \\ &= \sqrt{6^2 + (-7)^2} = \sqrt{85} = 9.22. \end{aligned}$$

$$\begin{aligned} RC &= \sqrt{(4+5)^2 + (-12+1)^2} \\ &= \sqrt{9^2 + (-11)^2} = \sqrt{202} \\ &= 14.21. \end{aligned}$$

If collinear, we would need $RB + BC = RC$ exactly.
 But $5 + \sqrt{85} \neq \sqrt{202}$ .

Hence, the given three points are not collinear.