Drag points on an axis to plot a linear graph (integer gradient and intercept only)
", "licence": "None specified"}, "statement": "Move the points $A$ and $B$ to make line $\\simplify{y={m}x+{c}}$
", "advice": "There are two ways to think about this, either by considering the points A and B, or by considering gradient ($m$) and intercept ($c$).
Let us firstly look at gradient and intercept.
$y=\\simplify{{m}x+{c}}$
Therefore $m = \\var{m}$ and $c = \\var{c}$.
The intercept tells us where the line intercepts the $y$-axis, so we can move one of the points to $(0,\\var{c})$ straight away.
The 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
$(1, \\var{c+m} )$.
The alternative method is to place the points directly
Choose any value of $x$ for your first point.
We will take $x = 1$ for this example.
Calculate the corresponding $y$-value by substituting into the equation $y=\\simplify{{m}x+{c}}$:
$y = \\var{m} \\times 1 + \\var{c}$
$y = \\var{m+c}$
\ntherefore the first point is $(1,\\var{m+c})$.
We repeat the process for $x=2$ (or any other value of $x$ that you choose)
$y = \\var{m} \\times 2 + \\var{c}$
\n$y = \\var{2*m+c}$
\ntherefore the second point is $(2,\\var{2*m+c})$.
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