Question

which points could be on the line that is parallel to \\( \overleftrightarrow{gh} \\) and passes through point j? check all that apply. \\( \square\\ (-3, 5) \\) \\( \square\\ (1, 5) \\) \\( \square\\ (3, -2) \\) \\( \square\\ (3, 2) \\) \\( \square\\ (5, 1) \\)

Explanation:

Step1: Find slope of \( \overleftrightarrow{GH} \)

Points \( G(-4,1) \) and \( H(2,-2) \). Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 1}{2 - (-4)} = \frac{-3}{6} = -\frac{1}{2} \).

Step2: Equation of line through \( J(1,3) \)

Using point - slope form \( y - y_1 = m(x - x_1) \), we get \( y - 3 = -\frac{1}{2}(x - 1) \), which simplifies to \( y = -\frac{1}{2}x+\frac{1}{2}+3=-\frac{1}{2}x+\frac{7}{2} \).

Step3: Check each point

• For \( (-3,5) \): Substitute \( x = - 3 \), \( y=-\frac{1}{2}(-3)+\frac{7}{2}=\frac{3 + 7}{2}=5 \). So, \( (-3,5) \) is on the line.
• For \( (1,5) \): Substitute \( x = 1 \), \( y=-\frac{1}{2}(1)+\frac{7}{2}=\frac{-1 + 7}{2}=3

eq5 \). Not on the line.

• For \( (3,-2) \): Substitute \( x = 3 \), \( y=-\frac{1}{2}(3)+\frac{7}{2}=\frac{-3 + 7}{2}=2

eq - 2 \). Not on the line.

• For \( (3,2) \): Substitute \( x = 3 \), \( y=-\frac{1}{2}(3)+\frac{7}{2}=\frac{-3 + 7}{2}=2 \). So, \( (3,2) \) is on the line.
• For \( (5,1) \): Substitute \( x = 5 \), \( y=-\frac{1}{2}(5)+\frac{7}{2}=\frac{-5 + 7}{2}=1 \). So, \( (5,1) \) is on the line.

Answer:

\( (-3,5) \), \( (3,2) \), \( (5,1) \)