Drag points on an axis to plot a linear graph (integer gradient and intercept only)
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There are two ways to think about this, either by considering two points, or by considering gradient ${m}$ and intercept ${c}$.
\nFirst let's consider choosing any two points:
\nChoose any value of ${x}$ for your first point.
\nWe will take ${x}=1$ for this example.
\nCalculate the corresponding ${y}$-value by substituting into the equation $\\simplify[!canonicalOrder]{y={m}x+{c}}$:
\n\\[y=(\\var{m}\\times 1)\\simplify{{+c}}\\]
\n\\[y=\\var{m}\\simplify{{+c}}\\]
therefore the first point is $(1,\\simplify{{m}+{c}})$.
We repeat the process for $x=2$ (or any other value of $x$ that you choose):
\n\\[y=(\\var{m}\\times 2)\\simplify{{+c}}\\]
\n\\[y=\\simplify{2{m}}\\simplify{{+c}}\\]
\ntherefore the second point is $(2,\\simplify{2{m}+{c}})$.
\nThe alternative method is to use the gradient and intercept:
\n$\\simplify[!canonicalOrder]{y={m}x+{c}}$
\nTherefore $\\simplify{m={m}}$ and $\\simplify{c={c}}$.
\nThe intercept tells us where the line intercepts the $y$-axis, so we can move one of the points to $(0,\\simplify{{c}})$ straight away.
\nThe gradient is a measure of 'how many units up for each unit across' (or 'units down' if the gradient is negative). In this case we want to go {uod} {abs(m)} for each unit across, so to place the second point we move across 1 unit on the $x$-axis and {uod} {abs(m)} on the $y$-axis, which takes us to $(1,\\simplify{{c+m}})$.
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