an equation of the line that contains the origin and the point (1,2) is ?

Answer 1

OK, so your line contains points (0,0) and (1,2).

As you go from (0,0) to (1,2),
your x goes up by 1 and your y goes up by 2.
Slope = rise/run = (change in y/change in x) = 2/1 = 2

The y-intercept is where the line crosses the y-axis.
That's where y is when x = 0.
Obviously, the y-intercept is 0.

In slope-intercept form: y = 2x + 0, or simply y = 2x.

Your equation is y=2x.

Hope that helps!

Answer 2

The equation of the line should be in the form y = mx + c where m is the gradient (steepness) of the line and c is the y-intercept (where the line crosses the y axis).

To work out the gradient of a line you have to look how far the line goes up for every unit it goes across. This line goes through (1,2) so when it has gone across to 1 from the origin it has gone 2 units up from the origin. Therefore the gradient (m) is 2.

As the line goes through the origin it crosses the Y axis (c) at 0.

So m = 2 and c = 0, this tells you that the equation of the line is y = 2x

Answer 3

(0,0) , (1,2) are the points. so the slope is (2-0)/(1-0) which is 2. so we get
y = 2x + b
then we fill the numbers in the equation.
2 = 2(1) + b
2 = 2+ b
b = 0 so the equation is simply y = 2x

Answer 4

The line is passing through the origin (0,0).
Hence, the equation of the line is of the form y = mx where m is the slope of the line.
The slope of the line is calculated as m = (y2-y1)/(x2-x1).
If (x1,y1) = (0,0) and (x2,y2) = (1,2), then substituting into the above formula, we get m = 2.

Hence, the equation of the line is y=2x.

Answer 5

y-y1=m(x-x1)
y1 and x1 are given points, so x1 will be 1 and y1 will be 2 since you were given the ordered pair (1,2)
m is the slope which you would get from the origin (0,0) and (1,2)
2-0/1-0=slope of 2

now just plug in what you know

y-2=2(x-1)

hope this helps! :)

Answer 6

f(x)=y
x=1 y=2 , and it's a function equations can't pass through points, equations are functions that have y=0