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{eqnline()}

The above graph shows a line which has an equation of the form $y=ax+b$ where $a$ and $b$ are integers.

You are given two points on the line as indicated on the diagram.

$(0,\\var{b}),\\;\\;(-1,\\var{b-a})$.

Your task is to input the equation of the line.

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Given a graph of a line of the form $y=ax+b$ where $a$ and $b$ are integers, find the equation of the line. The y-intercept is given.

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To find the gradient:

Form a triangle using the line as shown in the image above
The rise is how many units from the base of the triangle (perpendicular corner) up to the line along the y-axis
The run is how many units from the base of the triangle (perpendicular corner) to the line along the x-axis.

It is important to count from the perpendicular corner of the triangle as this affects whether the value is positive or negative depending on which direction the line is in. This in turn gives either a positive or negative gradient.

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Input the gradient of the line:

Gradient $=$[[0]]

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To find the line equation in the form $y=ax+b$, you need to know:

The gradient, $a$, as calculated above
The y-intercept, $b$, where the line crosses the $y$-axis

Substitute these values into the above equation to give the line equation of the graph.

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Hence, input the equation for the line in the diagram:

$y=\\;$[[0]]

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First Method.

We are given that the line goes through $(0,\\var{b})$ and $(-1,\\var{b-a})$ and the equation of the line is of the form $y=ax+b$

Hence:

1) At $x=0$ we have $y=\\var{b}$, and this gives $\\var{b}=a \\times 0 +b =b$ on putting $x=0$ into $y=ax+b$.
So $b=\\var{b}$.
2) At $x=-1$ we have $y=\\var{b-a}$, and this gives $\\var{b-a}=a \\times (-1) +b =\\simplify[all,!collectNumbers]{-a+{b}}$ on putting $x=-1$ into $y=ax+b$.
After rearranging we obtain $a=\\simplify[all,!collectNumbers]{{b}-{b-a}}=\\var{a}$.
So $a=\\var{a}$.
So the equation of the line is $\\simplify{y={a}*x+{b}}$.

Second Method.

The equation $y=ax+b$ tells us that the graph crosses the $y$-axis (when $x=0$) at $y=b$.
So looking at the graph we immediately see that $b=\\var{b}$.
$a$ is the gradient of the line and is given by the change from $(-1,\\var{b-a})$ to $(0,\\var{b})$:
\\[a=\\frac{\\text{Change in y}}{\\text{Change in x}}=\\frac{\\simplify[all,!collectNumbers]{({b-a}-{b})}}{-1-0}=\\var{a}\\]

For more help on line graphs click on this link.

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