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To find the [tex]\( y \)[/tex]-coordinate of the point that divides the directed line segment from [tex]\( J \)[/tex] to [tex]\( K \)[/tex] into a ratio of 2:3, we can use the section formula for dividing a line segment in a given ratio. Here, we are provided with the coordinates of points [tex]\( J \)[/tex] and [tex]\( K \)[/tex] and the ratio in which the segment is to be divided.
1. Identify the coordinates and ratio:
- Let [tex]\( J = (x_1, y_1) \)[/tex] and [tex]\( K = (x_2, y_2) \)[/tex].
- Given coordinates: [tex]\( J = (0, 0) \)[/tex] and [tex]\( K = (5, 5) \)[/tex].
- Given ratio: [tex]\( m:n = 2:3 \)[/tex].
2. Apply the section formula:
- The section formula states that the coordinates of a point [tex]\( P(x, y) \)[/tex] dividing a line segment [tex]\( AB \)[/tex] joining [tex]\( A(x_1, y_1) \)[/tex] and [tex]\( B(x_2, y_2) \)[/tex] in the ratio [tex]\( m:n \)[/tex] are given by:
[tex]\[ x = \frac{mx_2 + nx_1}{m+n} \quad \text{and} \quad y = \frac{my_2 + ny_1}{m+n} \][/tex]
3. Calculate the [tex]\( y \)[/tex]-coordinate using the formula:
- Given [tex]\( m = 2 \)[/tex], [tex]\( n = 3 \)[/tex], [tex]\( x_1 = 0 \)[/tex], [tex]\( y_1 = 0 \)[/tex], [tex]\( x_2 = 5 \)[/tex], and [tex]\( y_2 = 5 \)[/tex]:
[tex]\[ y = \frac{my_2 + ny_1}{m+n} = \frac{(2 \cdot 5) + (3 \cdot 0)}{2+3} \][/tex]
4. Substitute the values:
[tex]\[ y = \frac{(2 \cdot 5) + (3 \cdot 0)}{2+3} = \frac{10 + 0}{5} = \frac{10}{5} = 2 \][/tex]
Therefore, the [tex]\( y \)[/tex]-coordinate of the point that divides the directed line segment from [tex]\( J \)[/tex] to [tex]\( K \)[/tex] into a ratio of 2:3 is [tex]\( \boxed{2} \)[/tex].
1. Identify the coordinates and ratio:
- Let [tex]\( J = (x_1, y_1) \)[/tex] and [tex]\( K = (x_2, y_2) \)[/tex].
- Given coordinates: [tex]\( J = (0, 0) \)[/tex] and [tex]\( K = (5, 5) \)[/tex].
- Given ratio: [tex]\( m:n = 2:3 \)[/tex].
2. Apply the section formula:
- The section formula states that the coordinates of a point [tex]\( P(x, y) \)[/tex] dividing a line segment [tex]\( AB \)[/tex] joining [tex]\( A(x_1, y_1) \)[/tex] and [tex]\( B(x_2, y_2) \)[/tex] in the ratio [tex]\( m:n \)[/tex] are given by:
[tex]\[ x = \frac{mx_2 + nx_1}{m+n} \quad \text{and} \quad y = \frac{my_2 + ny_1}{m+n} \][/tex]
3. Calculate the [tex]\( y \)[/tex]-coordinate using the formula:
- Given [tex]\( m = 2 \)[/tex], [tex]\( n = 3 \)[/tex], [tex]\( x_1 = 0 \)[/tex], [tex]\( y_1 = 0 \)[/tex], [tex]\( x_2 = 5 \)[/tex], and [tex]\( y_2 = 5 \)[/tex]:
[tex]\[ y = \frac{my_2 + ny_1}{m+n} = \frac{(2 \cdot 5) + (3 \cdot 0)}{2+3} \][/tex]
4. Substitute the values:
[tex]\[ y = \frac{(2 \cdot 5) + (3 \cdot 0)}{2+3} = \frac{10 + 0}{5} = \frac{10}{5} = 2 \][/tex]
Therefore, the [tex]\( y \)[/tex]-coordinate of the point that divides the directed line segment from [tex]\( J \)[/tex] to [tex]\( K \)[/tex] into a ratio of 2:3 is [tex]\( \boxed{2} \)[/tex].
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