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To find the points of trisection of the line segment joining points [tex]\(A (2, -2)\)[/tex] and [tex]\(B (-7, 4)\)[/tex], we need to determine the coordinates of two points that divide the segment into three equal parts.
Let's denote the trisection points as [tex]\(P\)[/tex] and [tex]\(Q\)[/tex].
To find the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex], we can use the section formula. The section formula states that the coordinates of a point dividing a line segment in the ratio [tex]\(m:n\)[/tex] are given by:
[tex]\[ \left( \frac{m \cdot x_2 + n \cdot x_1}{m+n} , \frac{m \cdot y_2 + n \cdot y_1}{m+n} \right) \][/tex]
In our case:
- [tex]\(A (x_1, y_1) = (2, -2)\)[/tex]
- [tex]\(B (x_2, y_2) = (-7, 4)\)[/tex]
- We need to find the points of trisection, so the ratio [tex]\(m:n\)[/tex] will be [tex]\(1:2\)[/tex] for the first point [tex]\(P\)[/tex] and [tex]\(2:1\)[/tex] for the second point [tex]\(Q\)[/tex].
1. Finding [tex]\(P\)[/tex]:
For the first point [tex]\(P\)[/tex] which divides the segment in the ratio [tex]\(1:2\)[/tex]:
[tex]\[ P = \left( \frac{1 \cdot (-7) + 2 \cdot 2}{1+2} , \frac{1 \cdot 4 + 2 \cdot (-2)}{1+2} \right) \][/tex]
Calculating the coordinates:
[tex]\[ P_x = \frac{1 \cdot (-7) + 2 \cdot 2}{3} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1 \][/tex]
[tex]\[ P_y = \frac{1 \cdot 4 + 2 \cdot (-2)}{3} = \frac{4 - 4}{3} = \frac{0}{3} = 0 \][/tex]
Thus, [tex]\(P = (-1, 0)\)[/tex].
2. Finding [tex]\(Q\)[/tex]:
For the second point [tex]\(Q\)[/tex] which divides the segment in the ratio [tex]\(2:1\)[/tex]:
[tex]\[ Q = \left( \frac{2 \cdot (-7) + 1 \cdot 2}{2+1} , \frac{2 \cdot 4 + 1 \cdot (-2)}{2+1} \right) \][/tex]
Calculating the coordinates:
[tex]\[ Q_x = \frac{2 \cdot (-7) + 1 \cdot 2}{3} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4 \][/tex]
[tex]\[ Q_y = \frac{2 \cdot 4 + 1 \cdot (-2)}{3} = \frac{8 - 2}{3} = \frac{6}{3} = 2 \][/tex]
Thus, [tex]\(Q = (-4, 2)\)[/tex].
So, the coordinates of the points of trisection of the line segment joining [tex]\(A (2, -2)\)[/tex] and [tex]\(B (-7, 4)\)[/tex] are:
[tex]\[ (-1, 0) \quad \text{and} \quad (-4, 2) \][/tex]
Therefore, the correct answer is:
[tex]\[ \text{b. } (-1, 0), (-4, 2) \][/tex]
Let's denote the trisection points as [tex]\(P\)[/tex] and [tex]\(Q\)[/tex].
To find the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex], we can use the section formula. The section formula states that the coordinates of a point dividing a line segment in the ratio [tex]\(m:n\)[/tex] are given by:
[tex]\[ \left( \frac{m \cdot x_2 + n \cdot x_1}{m+n} , \frac{m \cdot y_2 + n \cdot y_1}{m+n} \right) \][/tex]
In our case:
- [tex]\(A (x_1, y_1) = (2, -2)\)[/tex]
- [tex]\(B (x_2, y_2) = (-7, 4)\)[/tex]
- We need to find the points of trisection, so the ratio [tex]\(m:n\)[/tex] will be [tex]\(1:2\)[/tex] for the first point [tex]\(P\)[/tex] and [tex]\(2:1\)[/tex] for the second point [tex]\(Q\)[/tex].
1. Finding [tex]\(P\)[/tex]:
For the first point [tex]\(P\)[/tex] which divides the segment in the ratio [tex]\(1:2\)[/tex]:
[tex]\[ P = \left( \frac{1 \cdot (-7) + 2 \cdot 2}{1+2} , \frac{1 \cdot 4 + 2 \cdot (-2)}{1+2} \right) \][/tex]
Calculating the coordinates:
[tex]\[ P_x = \frac{1 \cdot (-7) + 2 \cdot 2}{3} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1 \][/tex]
[tex]\[ P_y = \frac{1 \cdot 4 + 2 \cdot (-2)}{3} = \frac{4 - 4}{3} = \frac{0}{3} = 0 \][/tex]
Thus, [tex]\(P = (-1, 0)\)[/tex].
2. Finding [tex]\(Q\)[/tex]:
For the second point [tex]\(Q\)[/tex] which divides the segment in the ratio [tex]\(2:1\)[/tex]:
[tex]\[ Q = \left( \frac{2 \cdot (-7) + 1 \cdot 2}{2+1} , \frac{2 \cdot 4 + 1 \cdot (-2)}{2+1} \right) \][/tex]
Calculating the coordinates:
[tex]\[ Q_x = \frac{2 \cdot (-7) + 1 \cdot 2}{3} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4 \][/tex]
[tex]\[ Q_y = \frac{2 \cdot 4 + 1 \cdot (-2)}{3} = \frac{8 - 2}{3} = \frac{6}{3} = 2 \][/tex]
Thus, [tex]\(Q = (-4, 2)\)[/tex].
So, the coordinates of the points of trisection of the line segment joining [tex]\(A (2, -2)\)[/tex] and [tex]\(B (-7, 4)\)[/tex] are:
[tex]\[ (-1, 0) \quad \text{and} \quad (-4, 2) \][/tex]
Therefore, the correct answer is:
[tex]\[ \text{b. } (-1, 0), (-4, 2) \][/tex]
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