Question

identifying points on a parallel line
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 the slope of line \( \overleftrightarrow{GH} \)

The coordinates of \( G(-4,1) \) and \( H(2,-2) \).
The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
So, \( m_{GH}=\frac{-2 - 1}{2-(-4)}=\frac{-3}{6}=-\frac{1}{2} \).

Step2: Use the point - slope form to find the equation of the line passing through \( J(1,3) \)

The point - slope form is \( y - y_1=m(x - x_1) \).
Substitute \( m =-\frac{1}{2},x_1 = 1,y_1 = 3 \).
\( y-3=-\frac{1}{2}(x - 1) \), 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 \( x=-3 \) into \( y =-\frac{1}{2}x+\frac{7}{2} \), \( y=-\frac{1}{2}\times(-3)+\frac{7}{2}=\frac{3 + 7}{2}=5 \).

• For \( (1,5) \):

Substitute \( x = 1 \) into \( y=-\frac{1}{2}x+\frac{7}{2} \), \( y=-\frac{1}{2}\times1+\frac{7}{2}=\frac{-1 + 7}{2}=3
eq5 \).

• For \( (3,-2) \):

Substitute \( x = 3 \) into \( y=-\frac{1}{2}x+\frac{7}{2} \), \( y=-\frac{1}{2}\times3+\frac{7}{2}=\frac{-3 + 7}{2}=2
eq-2 \).

• For \( (3,2) \):

Substitute \( x = 3 \) into \( y=-\frac{1}{2}x+\frac{7}{2} \), \( y=-\frac{1}{2}\times3+\frac{7}{2}=\frac{-3 + 7}{2}=2 \).

• For \( (5,1) \):

Substitute \( x = 5 \) into \( y=-\frac{1}{2}x+\frac{7}{2} \), \( y=-\frac{1}{2}\times5+\frac{7}{2}=\frac{-5 + 7}{2}=1 \).

Answer:

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