Question

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

Explanation:

Step1: Find slope of line GH

Let \(G(-4,1)\) 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: Use point - slope form for line through J(1,3)

The point - slope form is \(y - y_1=m(x - x_1)\). Substituting \(m =-\frac{1}{2}\), \(x_1 = 1\), \(y_1=3\) gives \(y - 3=-\frac{1}{2}(x - 1)\), or \(y=-\frac{1}{2}x+\frac{1}{2}+3=-\frac{1}{2}x+\frac{7}{2}\).

Step3: Check each point

For \((-3,5)\): Substitute \(x=-3\) into \(y =-\frac{1}{2}x+\frac{7}{2}\), \(y=-\frac{1}{2}\times(-3)+\frac{7}{2}=\frac{3 + 7}{2}=5\), so \((-3,5)\) is on the line.
For \((1,5)\): Substitute \(x = 1\) into \(y=-\frac{1}{2}x+\frac{7}{2}\), \(y=-\frac{1}{2}\times1+\frac{7}{2}=3
eq5\), so \((1,5)\) is not on the line.
For \((3,-2)\): Substitute \(x = 3\) into \(y=-\frac{1}{2}x+\frac{7}{2}\), \(y=-\frac{1}{2}\times3+\frac{7}{2}=2
eq - 2\), so \((3,-2)\) is not on the line.
For \((3,2)\): Substitute \(x = 3\) into \(y=-\frac{1}{2}x+\frac{7}{2}\), \(y=-\frac{1}{2}\times3+\frac{7}{2}=2\), so \((3,2)\) is on the line.
For \((5,1)\): Substitute \(x = 5\) into \(y=-\frac{1}{2}x+\frac{7}{2}\), \(y=-\frac{1}{2}\times5+\frac{7}{2}=1\), so \((5,1)\) is on the line.

Answer:

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