Question

1. point g lies on ef such that eg:fg is 1:3. graph g.

Explanation:

Step1: Identify coordinates of E and F

From the graph, \( E(-2, -9) \) and \( F(6, -1) \).

Step2: Use section formula (internal division)

The ratio \( m:n = 1:3 \). The section formula for a point \( G(x,y) \) dividing \( EF \) in ratio \( m:n \) is \( x=\frac{mx_2 + nx_1}{m + n} \), \( y=\frac{my_2 + ny_1}{m + n} \).
Substitute \( m = 1 \), \( n = 3 \), \( x_1=-2 \), \( x_2 = 6 \), \( y_1=-9 \), \( y_2=-1 \).
\( x=\frac{1\times6 + 3\times(-2)}{1 + 3}=\frac{6 - 6}{4}=0 \)
\( y=\frac{1\times(-1)+3\times(-9)}{1 + 3}=\frac{-1 - 27}{4}=\frac{-28}{4}=-7 \)

Answer:

The coordinates of \( G \) are \( (0, -7) \), and it can be graphed at the point where \( x = 0 \) and \( y=-7 \) on the coordinate plane.