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First Method.

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You 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 $y=ax+b$

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Hence:

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1) At $x=0$ we have $y=\\var{b}$, and this gives $\\var{b}=a \\times 0 +b =b$ on putting $x=0$ into $y=ax+b$.

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So $b=\\var{b}$.

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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 $y=ax+b$.

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On rearranging we obtain $a=\\simplify[all,!collectNumbers]{{b}-{b-a}}=\\var{a}$.

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So $a=\\var{a}$.

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So the equation of the line is $\\simplify{y={a}*x+{b}}$.

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Second Method.

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The equation $y=ax+b$ tells us that the graph crosses the $y$-axis (when $x=0$) at $y=b$.

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So looking at the graph we immediately see that $b=\\var{b}$.

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$a$ is the gradient of the line and is given by the change from $(-1,\\var{b-a})$ to $(0,\\var{b})$:

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\$a=\\frac{\\text{Change in y}}{\\text{Change in x}}=\\frac{\\simplify[all,!collectNumbers]{({b-a}-{b})}}{-1-0}=\\var{a}\$

There are copious comments in the definition of the function eqnline about the voodoo needed to have a JSXGraph diagram interact with the input box for a part.

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Write the equation of the line in the diagram. The line described by your equation will also be drawn on the diagram.

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$y=\\;$[]

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\n\n\n", "language": "javascript"}}, "statement": "

{eqnline(a,b,x2,y2)}

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

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You are given two points on the line, $(0,\\var{b})$ and $(\\var{x2},\\var{y2})$, as indicated on the diagram.

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