Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
Step-by-Step Solution
Key Concept: Use the section formula for internal division of a line segment. If a point P divides AB in the ratio m:n (from A to P is m parts and from P to B is n parts), then \(P\bigl(\frac{n x_1 + m x_2}{m+n},\;\frac{n y_1 + m y_2}{m+n}\bigr)\). For trisection, the ratios are 1:2 and 2:1.
1. Identify the end points\
\[A( x_1 , y_1 ) = (2, -2), \quad B( x_2 , y_2 ) = (-7, 4)\]
2. First point of trisection – it divides AB in the ratio \(1:2\) (one part from A, two parts from the point to B).\
Using the section formula:\
\[P_1\bigl(\frac{2\cdot x_1 + 1\cdot x_2}{1+2},\;\frac{2\cdot y_1 + 1\cdot y_2}{1+2}\bigr)\]
Substitute the coordinates:\
\[P_1\bigl(\frac{2\cdot 2 + 1\cdot (-7)}{3},\;\frac{2\cdot (-2) + 1\cdot 4}{3}\bigr)\]
\[P_1\bigl(\frac{4-7}{3},\;\frac{-4+4}{3}\bigr) = \bigl(-1,\;0\bigr)\]
3. Second point of trisection – it divides AB in the ratio \(2:1\) (two parts from A, one part from the point to B).\
Using the section formula again:\
\[P_2\bigl(\frac{1\cdot x_1 + 2\cdot x_2}{2+1},\;\frac{1\cdot y_1 + 2\cdot y_2}{2+1}\bigr)\]
Substitute the coordinates:\
\[P_2\bigl(\frac{1\cdot 2 + 2\cdot (-7)}{3},\;\frac{1\cdot (-2) + 2\cdot 4}{3}\bigr)\]
\[P_2\bigl(\frac{2-14}{3},\;\frac{-2+8}{3}\bigr) = \bigl(-4,\;2\bigr)\]
4. Result – The two points that trisect the segment AB are \((-1,0)\) and \((-4,2)\).
Correct Answer: The points of trisection are \((-1,0)\) and \((-4,2)\).