find an equation of a line that contains the origin and the given point (2,-2)?

Answer 1

You are given two points, (0,0) and (2,-2). With these two points you can calculate the slope of the equation.

m = (y1-y2)/(x1-x2) = [0-(-2)]/[0-2] = 2/-2 = -1

Now that we have the slope we can use the standard equation with one of the points.

y - y1 = m(x - x1)

The best point to use is (0,0) because it cancels out and makes the work easier.

y - 0 = -1(x - 0)

y = -x

Answer 2

The origin is (0,0) and the point you need is (2, -2)

The equation of a line is y=mx+b

Where m is the slope, and b is the y intercept.

m = (y2 - y1) / (x2-x1) = -2-0 / 2-0 = -2/2 = -1

y = -x + b

Now, to find the y intercept just set y=0. Now, since we know it passes through the origin, 0,0, we can say that when y=0, x=0, therefore, b is 0

y = -x

Check:

When y=-2, x is -(-2) or 2.

So this works.

Answer 3

Well the equation of a line is y = mx + b where m is the slope of the line and b is the y intercept. Since it goes through the origin, b = 0. Now to get the slope, just plug in y, x and b into the equation and solve for m.

-2 = m*2 + 0
m = -1

So to get the equation of the line plug in m and b into the equation of the line. So in this case you have y = -x as your equation.

Answer 4

first find the slope (m)
m=(y2-y1)/(x2-x1)
m=(-2-0)/(2-0) or -2/2 or -1 (all are the same thing)

then use slope intercept to find out the equation. y=mx+b
-2=-1(2) + b
-2=-2+b
0=b

so the equation would be y=-x +0. then put that in standard form.
so your final answer is x+y=0