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

Apr 28, 2017

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?

Apr 28, 2017

$y = x$

The y-intercept $\to \left(x , y\right) = \left(0 , 0\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Initial reaction}}$

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 ${y}_{g} = f \left({x}_{g}\right) \leftarrow \text{ some function of } x$

Under these conditions we can write:
Let the gradient be $m = \frac{{y}_{g} - 1}{{x}_{g} - 1}$

Let the y-intercept be $c$

Then we have $y = m x + c$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{However}}$

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

Set the point as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(1 , 1\right)$

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

Thus the standard form of $y = m x + c$ can only work with the following two conditions:

$m = 1 \mathmr{and} c = 0$

Note that $c$ is the y-intercept

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

$y = 1 \left(0\right) + 0$

$\implies y = 0 \to {P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(0 , 0\right)$

Thus $y = m x + c \text{ becomes } y = x$