Question

How do you find a standard form equation for the line with point (1, 1) and has a y-intercept?

Answer 1

see explanation

Explanation:

Standard form of an equation involves two intercepts, x and y intercepts...

...but you only have one intercept, and it's not complete.
Is this all the information you have?

Answer 2

`y=x`

The y-intercept `->(x,y)=(0,0)`

Explanation:

`color(blue)("Initial reaction")`

Initially, due to the lack of information it is tempting to write:

The gradient is such that you read left to right on the x-axis> So to force this to work we must have:

For the gradient set `x_g>1`
For the gradient set `y_g>1` where `y_g=f(x_g) larr" some function of "x`

Under these conditions we can write:
Let the gradient be `m=(y_g-1)/(x_g-1)`

Let the y-intercept be `c`

Then we have `y=mx+c`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("However")`

We are given that the line passes through the point (1,1)

Set the point as `P_1->(x_1,y_1)=(1,1)`

So we have the answer of 1 when `x=1`

Thus the standard form of `y=mx+c` can only work with the following two conditions:

`m=1 and c=0`

Note that `c` is the y-intercept

Thus as the y-intercept is at `x=0` we have:

`y=1(0)+0`

`=>y=0->P_2->(x_2,y_2)=(0,0)`

Thus `y=mx+c" becomes "y=x`