What is an equation of the line that passes
thrcugh the point (-2,5) and is parallel to the
line x - 2y = 6?

So you have y-y1=m(x-x1) and you are told that the slope has to be parallel to the line x-2y=6

1.) solve x-2y=6 for y

x-2y=6

Subtract x from both sides to get

-2y=6-x

Divide each side by -2 to get

y=-3-(x/2)

Now the slope for this equation is -1/2 and the line has to be parallel to this equation. Therefore the slope of the line that is parallel to this equation is going to be -1/2.

Now You have to use the point (2,-5) as your x1 and y1 and the slope that is parallel to the line x-2y=6, which is -1/2

y-(-5)=-1/2[x-(-2)]

Simplifying this you get

y+5=-1/2(x+2)

Above is how you write the equation in point slope form.

Now if it is asking you for slope-intercept form all you do is distribute -1/2 and subtract 5

y+5=-1/2(x+2)

y+5=-1x/2-1
Subtracting 5 from both sides gives you

y=-x/2 +4. Hope this helps. Let me know if you have questions