What is an equation of the line that passes through the point (2,-5) and is parallel to the line x-2y=16?

Answer 1

Parallel lines have the same slope, so we need to find the slope of the given line. Let's put it in slope-intercept form (y=mx+b):

x - 2y = 16

Subtract x from both sides:

-2y = -x+16

Divide each term by (-2):

y=(1/2)x - 8

Comparing this to y = mx + b, where m is the slope, and we see m = 1/2.

Plug the given point in for x and y to get the value of b:

-5 = (1/2)(2) + b

Multiply:

-5 = 1 + b

Subtract 1 from both sides to get b:

-6 = b.

Plug m and b into the equation for a line:

y = (1/2)x - 6

I hope this helps, and let me know if you have any questions!

Have fun mathing!

Liz Z.

Answer 2

You may do this (and many teachers and tutors will teach you to do it), but please do it once and forever:

Convert the equation from standard form (Ax+By=C) to slope-intercept form (y=mx+b) and use the SAME SLOPE for the parallel line.

It makes sense to convert Ax+By=C to y=mx+b and remember that the slope: m = (-A/B).

The parallel line has the equation x-2y=C, so let's determine C by using the point (2,-5).

2 - 2(-5) = C

12 = C

The equation of the parallel line (in Standard Form) is: x - 2y = 12

Answer 3

x-2y=16

2y=x-16

y = (1/2)x - 8 which has slope 1/2, and any parallel line also has slope = 1/2

Use the point slope formula, plug in (2,-5) for (h,k) and 1/2 = m

y=m(x-k)+h

y=(1/2)(x-2)-5 = x/2 -6

2y=x-12

x-2y=12