Certainly! Let's solve the problem step-by-step.
We are given two points [tex]\( J \)[/tex] and [tex]\( K \)[/tex] with their coordinates, and we need to find the [tex]\( y \)[/tex]-coordinate of the point that divides the directed line segment from [tex]\( J \)[/tex] to [tex]\( K \)[/tex] in the ratio [tex]\(2:3\)[/tex].
The coordinates of the points [tex]\( J \)[/tex] and [tex]\( K \)[/tex] are:
[tex]\[ J = (-6, 5) \][/tex]
[tex]\[ K = (7, -5) \][/tex]
The ratio is given as [tex]\( m:n = 2:3 \)[/tex]. This means [tex]\( m = 2 \)[/tex] and [tex]\( n = 3 \)[/tex].
We need to find the y-coordinate of the point that divides the segment in this ratio. The formula for finding the [tex]\( y \)[/tex]-coordinate [tex]\( y \)[/tex] of the dividing point is given by:
[tex]\[ y = \left(\frac{m}{m+n}\right) \left(y_K - y_J\right) + y_J \][/tex]
Here:
- [tex]\( y_J \)[/tex] is the [tex]\( y \)[/tex]-coordinate of point [tex]\( J \)[/tex]
- [tex]\( y_K \)[/tex] is the [tex]\( y \)[/tex]-coordinate of point [tex]\( K \)[/tex]
- [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are the parts of the given ratio
Plugging in the values we have:
[tex]\[ y_J = 5 \][/tex]
[tex]\[ y_K = -5 \][/tex]
[tex]\[ m = 2 \][/tex]
[tex]\[ n = 3 \][/tex]
Firstly, compute the sum [tex]\( m + n \)[/tex]:
[tex]\[ m + n = 2 + 3 = 5 \][/tex]
Then, calculate the fraction [tex]\(\frac{m}{m+n}\)[/tex]:
[tex]\[ \frac{m}{m+n} = \frac{2}{5} \][/tex]
Now, find the difference [tex]\( y_K - y_J \)[/tex]:
[tex]\[ y_K - y_J = -5 - 5 = -10 \][/tex]
Now we apply these values into our formula:
[tex]\[ y = \left(\frac{2}{5}\right) (-10) + 5 \][/tex]
First, compute [tex]\( \left(\frac{2}{5}\right) (-10) \)[/tex]:
[tex]\[ \left(\frac{2}{5}\right) (-10) = \frac{2 \times -10}{5} = \frac{-20}{5} = -4 \][/tex]
Now add [tex]\( y_J \)[/tex] to this value:
[tex]\[ y = -4 + 5 = 1 \][/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 [tex]\( 2:3 \)[/tex] is:
[tex]\[ \boxed{1.0} \][/tex]
