Which points could be on the line that is parallel toand passes through point J? Check all that apply.(-3,5)(1,5)(3,-2)(3, 2)(5,1)

Answer:(-3, 5), (3, 2), (5, 1)Step-by-step explanation:Parallel lines have the same slope.The formula of a slope:[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]Substitute the coordinates of the given points G(-4, 1) and H(2, -2):[tex]m=\dfrac{-2-1}{2-(-4)}=\dfrac{-3}{6}=-\dfrac{1}{2}[/tex]J(1, 3). Let other point (x, y).Substitute to the slope:[tex]\dfrac{y-3}{x-1}=\dfrac{-1}{2}[/tex]            cross multiply[tex]2(y-3)=-1(x-1)[/tex]    use the distributive property[tex]2y+(2)(-3)=-x+(-1)(-1)[/tex][tex]2y-6=-x+1[/tex]        add 6 to both sides[tex]2y=-x+7[/tex]      divide both sides by 2[tex]y=-\dfrac{1}{2}x+\dfrac{7}{2}[/tex]Check the equality for coordinates of each point:[tex](-3, 5)\\\\5=-\dfrac{1}{2}(-3)+\dfrac{7}{2}\\\\5=\dfrac{3}{2}+\dfrac{7}{2}\\\\5=\dfrac{10}{2}\\\\5=5\qquad\bold{CORRECT}[/tex][tex](1,\ 5)\\\\5=-\dfrac{1}{2}(1)+\dfrac{7}{2}\\\\5=-\dfrac{1}{2}+\dfrac{7}{2}\\\\5=\dfrac{6}{2}\\\\5=3\qquad\bold{FALSE}[/tex][tex](3,\ -2)\\\\-2=-\dfrac{1}{2}(3)+\dfrac{7}{2}\\\\-2=-\dfrac{3}{2}+\dfrac{7}{2}\\\\-2=\dfrac{4}{2}\\\\-2=2\qquad\bold{FALSE}[/tex][tex](3,\ 2)\\\\2=-\dfrac{1}{2}(3)+\dfrac{7}{2}\\\\2=-\dfrac{3}{2}+\dfrac{7}{2}\\\\2=\dfrac{4}{2}\\\\2=2\qquad\bold{CORRECT}[/tex][tex](5,\ 1)\\\\1=-\dfrac{1}{2}(5)+\dfrac{7}{2}\\\\1=-\dfrac{5}{2}+\dfrac{7}{2}\\\\1=\dfrac{2}{2}\\\\1=1\qquad\bold{CORRECT}[/tex]