Answer :
To find the coordinates of point P on the line segment AB based on the given ratios, we can use the section formula. The section formula helps find a point that divides a line segment internally in a given ratio. If a point P divides the line segment AB in the ratio [tex]m:n[/tex], then the coordinates of P are given by:
[tex]P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)[/tex]
where [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] are the coordinates of points A and B respectively.
Let's solve each part using this formula:
(a) A(0, 1), B(15, 7) and AP : PB = 1 : 2
[tex]m = 1[/tex], [tex]n = 2[/tex]
[tex]x = \frac{1 \cdot 15 + 2 \cdot 0}{1 + 2} = \frac{15 + 0}{3} = 5[/tex]
[tex]y = \frac{1 \cdot 7 + 2 \cdot 1}{1 + 2} = \frac{7 + 2}{3} = 3[/tex]
So, the coordinates of P are [tex](5, 3)[/tex].
(b) A(7, 10), B(-5, 6) and AP : PB = 1 : 3
[tex]m = 1[/tex], [tex]n = 3[/tex]
[tex]x = \frac{1 \cdot (-5) + 3 \cdot 7}{1 + 3} = \frac{-5 + 21}{4} = 4[/tex]
[tex]y = \frac{1 \cdot 6 + 3 \cdot 10}{1 + 3} = \frac{6 + 30}{4} = 9[/tex]
So, the coordinates of P are [tex](4, 9)[/tex].
(c) A(1, 8), B(-4, 3) and AP : PB = 4 : 1
[tex]m = 4[/tex], [tex]n = 1[/tex]
[tex]x = \frac{4 \cdot (-4) + 1 \cdot 1}{4 + 1} = \frac{-16 + 1}{5} = -3[/tex]
[tex]y = \frac{4 \cdot 3 + 1 \cdot 8}{4 + 1} = \frac{12 + 8}{5} = 4[/tex]
So, the coordinates of P are [tex](-3, 4)[/tex].
(d) A(-8, -9), B(8, -1) and AP : PB = 3 : 5
[tex]m = 3[/tex], [tex]n = 5[/tex]
[tex]x = \frac{3 \cdot 8 + 5 \cdot (-8)}{3 + 5} = \frac{24 - 40}{8} = -2[/tex]
[tex]y = \frac{3 \cdot (-1) + 5 \cdot (-9)}{3 + 5} = \frac{-3 - 45}{8} = -6[/tex]
So, the coordinates of P are [tex](-2, -6)[/tex].
Each result tells us where P is located on the line segment AB according to the respective ratio given in the problem.
