Find the y value for point F such that DF and EF form a 1:3 ratio. (5 points) Segment DE is shown. D is at negative 3, negative 3. E is at 2, 4. 1.67 −1.33 0.75 −1.25
Answer:
The correct option is 4.
Step-by-step explanation:
It is given that DF and EF form a 1:3 ratio. The coordinates of the end points of the segment DE are D(-3,-3) and E(2,4).
According to the section formula, if a point p divides a line in m:n, then the coordinates of p are
\((\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})\)
y value for point F is calculated as
\(y=\frac{(1)y_2+(3)y_1}{1+3}\)
The end points of the segment DE are D(-3,-3) and E(2,4).
\(y=\frac{(1)(4)+(3)(-3)}{4}\)
\(y=\frac{4-9}{4}\)
\(y=\frac{-5}{4}\)
\(y=-1.25\)
The y- value for point F is -1.25. Therefore the correct option is 4.