Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Parallel lines have the same slope.

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute the coordinates of the given points G(-4, 1) and H(2, -2):

$m=\dfrac{-2-1}{2-(-4)}=\dfrac{-3}{6}=-\dfrac{1}{2}$

J(1, 3). Let other point (x, y).

Substitute to the slope:

$\dfrac{y-3}{x-1}=\dfrac{-1}{2}$            cross multiply

$2(y-3)=-1(x-1)$    use the distributive property

$2y+(2)(-3)=-x+(-1)(-1)$

$2y-6=-x+1$        add 6 to both sides

$2y=-x+7$      divide both sides by 2

$y=-\dfrac{1}{2}x+\dfrac{7}{2}$

Check the equality for coordinates of each point:

$(-3, 5)\\\\5=-\dfrac{1}{2}(-3)+\dfrac{7}{2}\\\\5=\dfrac{3}{2}+\dfrac{7}{2}\\\\5=\dfrac{10}{2}\\\\5=5\qquad\bold{CORRECT}$

$(1,\ 5)\\\\5=-\dfrac{1}{2}(1)+\dfrac{7}{2}\\\\5=-\dfrac{1}{2}+\dfrac{7}{2}\\\\5=\dfrac{6}{2}\\\\5=3\qquad\bold{FALSE}$

$(3,\ -2)\\\\-2=-\dfrac{1}{2}(3)+\dfrac{7}{2}\\\\-2=-\dfrac{3}{2}+\dfrac{7}{2}\\\\-2=\dfrac{4}{2}\\\\-2=2\qquad\bold{FALSE}$

$(3,\ 2)\\\\2=-\dfrac{1}{2}(3)+\dfrac{7}{2}\\\\2=-\dfrac{3}{2}+\dfrac{7}{2}\\\\2=\dfrac{4}{2}\\\\2=2\qquad\bold{CORRECT}$

$(5,\ 1)\\\\1=-\dfrac{1}{2}(5)+\dfrac{7}{2}\\\\1=-\dfrac{5}{2}+\dfrac{7}{2}\\\\1=\dfrac{2}{2}\\\\1=1\qquad\bold{CORRECT}$