Points \(A(-6,10),\ B(-4,6),\ C(3,-8)\) are collinear such that \(AB=\dfrac{2}{9}\,AC\).
True.
Step 1: Recall the distance formula.
The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 2: Find the distance AB.
Points are \(A(-6,10)\) and \(B(-4,6)\).
\[ AB = \sqrt{((-4)-(-6))^2 + (6-10)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
Step 3: Find the distance AC.
Points are \(A(-6,10)\) and \(C(3,-8)\).
\[ AC = \sqrt{((3)-(-6))^2 + (-8-10)^2} = \sqrt{(9)^2 + (-18)^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5} \]
Step 4: Compare the ratio \(\dfrac{AB}{AC}\).
\[ \dfrac{AB}{AC} = \dfrac{2\sqrt{5}}{9\sqrt{5}} = \dfrac{2}{9} \]
This matches the condition given in the question.
Step 5: Check collinearity using area of triangle.
Three points are collinear if the area of the triangle formed by them is 0.
For \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\):
\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\).
Substitute \(A(-6,10)\), \(B(-4,6)\), \(C(3,-8)\):
\(\text{Area} = \dfrac{1}{2}\left|(-6)(6-(-8)) + (-4)((-8)-10) + 3(10-6)\right|\)
Now compute each term:
Add them: \(-84 + 72 + 12 = 0\)
So, \(\text{Area} = \dfrac{1}{2}|0| = 0\).
Step 6: Final Conclusion.
Since the area is 0, points A, B, and C are collinear, and from Step 4 we also have \(AB = \tfrac{2}{9}AC\). Hence, the statement is True.